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Construct Tree from given Inorder and Preorder traversals

  • Difficulty Level : Hard
  • Last Updated : 26 Nov, 2021

Let us consider the below traversals:
Inorder sequence: D B E A F C 
Preorder sequence: A B D E C F

In a Preorder sequence, the leftmost element is the root of the tree. So we know ‘A’ is the root for given sequences. By searching ‘A’ in the Inorder sequence, we can find out all elements on the left side of ‘A’ is in the left subtree, and elements on right in the right subtree. So we know the below structure now. 

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                 A
               /   \
             /       \
           D B E     F C

We recursively follow the above steps and get the following tree.

         A
       /   \
     /       \
    B         C
   / \        /
 /     \    /
D       E  F

Algorithm: buildTree() 
1) Pick an element from Preorder. Increment a Preorder Index Variable (preIndex in below code) to pick the next element in the next recursive call. 
2) Create a new tree node tNode with the data as the picked element. 
3) Find the picked element’s index in Inorder. Let the index be inIndex. 
4) Call buildTree for elements before inIndex and make the built tree as a left subtree of tNode. 
5) Call buildTree for elements after inIndex and make the built tree as a right subtree of tNode. 
6) return tNode.



C++




/* C++ program to construct tree using
inorder and preorder traversals */
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
class node
{
    public:
    char data;
    node* left;
    node* right;
};
 
/* Prototypes for utility functions */
int search(char arr[], int strt, int end, char value);
node* newNode(char data);
 
/* Recursive function to construct binary
of size len from Inorder traversal in[]
and Preorder traversal pre[]. Initial values
of inStrt and inEnd should be 0 and len -1.
The function doesn't do any error checking
for cases where inorder and preorder do not
form a tree */
node* buildTree(char in[], char pre[], int inStrt, int inEnd)
{
    static int preIndex = 0;
 
    if (inStrt > inEnd)
        return NULL;
 
    /* Pick current node from Preorder
    traversal using preIndex
    and increment preIndex */
    node* tNode = newNode(pre[preIndex++]);
 
    /* If this node has no children then return */
    if (inStrt == inEnd)
        return tNode;
 
    /* Else find the index of this node in Inorder traversal */
    int inIndex = search(in, inStrt, inEnd, tNode->data);
 
    /* Using index in Inorder traversal, construct left and
    right subtress */
    tNode->left = buildTree(in, pre, inStrt, inIndex - 1);
    tNode->right = buildTree(in, pre, inIndex + 1, inEnd);
 
    return tNode;
}
 
/* UTILITY FUNCTIONS */
/* Function to find index of value in arr[start...end]
The function assumes that value is present in in[] */
int search(char arr[], int strt, int end, char value)
{
    int i;
    for (i = strt; i <= end; i++)
    {
        if (arr[i] == value)
            return i;
    }
}
 
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
node* newNode(char data)
{
    node* Node = new node();
    Node->data = data;
    Node->left = NULL;
    Node->right = NULL;
 
    return (Node);
}
 
/* This function is here just to test buildTree() */
void printInorder(node* node)
{
    if (node == NULL)
        return;
 
    /* first recur on left child */
    printInorder(node->left);
 
    /* then print the data of node */
    cout<<node->data<<" ";
 
    /* now recur on right child */
    printInorder(node->right);
}
 
/* Driver code */
int main()
{
    char in[] = { 'D', 'B', 'E', 'A', 'F', 'C' };
    char pre[] = { 'A', 'B', 'D', 'E', 'C', 'F' };
    int len = sizeof(in) / sizeof(in[0]);
    node* root = buildTree(in, pre, 0, len - 1);
 
    /* Let us test the built tree by
    printing Inorder traversal */
    cout << "Inorder traversal of the constructed tree is \n";
    printInorder(root);
}
 
// This is code is contributed by rathbhupendra


C




/* program to construct tree using inorder and preorder traversals */
#include <stdio.h>
#include <stdlib.h>
 
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node {
    char data;
    struct node* left;
    struct node* right;
};
 
/* Prototypes for utility functions */
int search(char arr[], int strt, int end, char value);
struct node* newNode(char data);
 
/* Recursive function to construct binary of size len from
   Inorder traversal in[] and Preorder traversal pre[].  Initial values
   of inStrt and inEnd should be 0 and len -1.  The function doesn't
   do any error checking for cases where inorder and preorder
   do not form a tree */
struct node* buildTree(char in[], char pre[], int inStrt, int inEnd)
{
    static int preIndex = 0;
 
    if (inStrt > inEnd)
        return NULL;
 
    /* Pick current node from Preorder traversal using preIndex
    and increment preIndex */
    struct node* tNode = newNode(pre[preIndex++]);
 
    /* If this node has no children then return */
    if (inStrt == inEnd)
        return tNode;
 
    /* Else find the index of this node in Inorder traversal */
    int inIndex = search(in, inStrt, inEnd, tNode->data);
 
    /* Using index in Inorder traversal, construct left and
     right subtress */
    tNode->left = buildTree(in, pre, inStrt, inIndex - 1);
    tNode->right = buildTree(in, pre, inIndex + 1, inEnd);
 
    return tNode;
}
 
/* UTILITY FUNCTIONS */
/* Function to find index of value in arr[start...end]
   The function assumes that value is present in in[] */
int search(char arr[], int strt, int end, char value)
{
    int i;
    for (i = strt; i <= end; i++) {
        if (arr[i] == value)
            return i;
    }
}
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct node* newNode(char data)
{
    struct node* node = (struct node*)malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
 
    return (node);
}
 
/* This function is here just to test buildTree() */
void printInorder(struct node* node)
{
    if (node == NULL)
        return;
 
    /* first recur on left child */
    printInorder(node->left);
 
    /* then print the data of node */
    printf("%c ", node->data);
 
    /* now recur on right child */
    printInorder(node->right);
}
 
/* Driver program to test above functions */
int main()
{
    char in[] = { 'D', 'B', 'E', 'A', 'F', 'C' };
    char pre[] = { 'A', 'B', 'D', 'E', 'C', 'F' };
    int len = sizeof(in) / sizeof(in[0]);
    struct node* root = buildTree(in, pre, 0, len - 1);
 
    /* Let us test the built tree by printing Inorder traversal */
    printf("Inorder traversal of the constructed tree is \n");
    printInorder(root);
    getchar();
}


Java




// Java program to construct a tree using inorder and preorder traversal
 
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
class Node {
    char data;
    Node left, right;
 
    Node(char item)
    {
        data = item;
        left = right = null;
    }
}
 
class BinaryTree {
    Node root;
    static int preIndex = 0;
 
    /* Recursive function to construct binary of size len from
       Inorder traversal in[] and Preorder traversal pre[].
       Initial values of inStrt and inEnd should be 0 and len -1. 
       The function doesn't do any error checking for cases where
       inorder and preorder do not form a tree */
    Node buildTree(char in[], char pre[], int inStrt, int inEnd)
    {
        if (inStrt > inEnd)
            return null;
 
        /* Pick current node from Preorder traversal using preIndex
           and increment preIndex */
        Node tNode = new Node(pre[preIndex++]);
 
        /* If this node has no children then return */
        if (inStrt == inEnd)
            return tNode;
 
        /* Else find the index of this node in Inorder traversal */
        int inIndex = search(in, inStrt, inEnd, tNode.data);
 
        /* Using index in Inorder traversal, construct left and
           right subtress */
        tNode.left = buildTree(in, pre, inStrt, inIndex - 1);
        tNode.right = buildTree(in, pre, inIndex + 1, inEnd);
 
        return tNode;
    }
 
    /* UTILITY FUNCTIONS */
 
    /* Function to find index of value in arr[start...end]
     The function assumes that value is present in in[] */
    int search(char arr[], int strt, int end, char value)
    {
        int i;
        for (i = strt; i <= end; i++) {
            if (arr[i] == value)
                return i;
        }
        return i;
    }
 
    /* This function is here just to test buildTree() */
    void printInorder(Node node)
    {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        System.out.print(node.data + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    // driver program to test above functions
    public static void main(String args[])
    {
        BinaryTree tree = new BinaryTree();
        char in[] = new char[] { 'D', 'B', 'E', 'A', 'F', 'C' };
        char pre[] = new char[] { 'A', 'B', 'D', 'E', 'C', 'F' };
        int len = in.length;
        Node root = tree.buildTree(in, pre, 0, len - 1);
 
        // building the tree by printing inorder traversal
        System.out.println("Inorder traversal of constructed tree is : ");
        tree.printInorder(root);
    }
}
 
// This code has been contributed by Mayank Jaiswal


Python




# Python program to construct tree using inorder and
# preorder traversals
 
# A binary tree node
class Node:
     
    # Constructor to create a new node
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
"""Recursive function to construct binary of size len from
   Inorder traversal in[] and Preorder traversal pre[].  Initial values
   of inStrt and inEnd should be 0 and len -1.  The function doesn't
   do any error checking for cases where inorder and preorder
   do not form a tree """
def buildTree(inOrder, preOrder, inStrt, inEnd):
     
    if (inStrt > inEnd):
        return None
 
    # Pich current node from Preorder traversal using
    # preIndex and increment preIndex
    tNode = Node(preOrder[buildTree.preIndex])
    buildTree.preIndex += 1
 
    # If this node has no children then return
    if inStrt == inEnd :
        return tNode
 
    # Else find the index of this node in Inorder traversal
    inIndex = search(inOrder, inStrt, inEnd, tNode.data)
     
    # Using index in Inorder Traversal, construct left
    # and right subtrees
    tNode.left = buildTree(inOrder, preOrder, inStrt, inIndex-1)
    tNode.right = buildTree(inOrder, preOrder, inIndex + 1, inEnd)
 
    return tNode
 
# UTILITY FUNCTIONS
# Function to find index of value in arr[start...end]
# The function assumes that value is present in inOrder[]
 
def search(arr, start, end, value):
    for i in range(start, end + 1):
        if arr[i] == value:
            return i
 
def printInorder(node):
    if node is None:
        return
     
    # first recur on left child
    printInorder(node.left)
     
    # then print the data of node
    print node.data,
 
    # now recur on right child
    printInorder(node.right)
     
# Driver program to test above function
inOrder = ['D', 'B', 'E', 'A', 'F', 'C']
preOrder = ['A', 'B', 'D', 'E', 'C', 'F']
# Static variable preIndex
buildTree.preIndex = 0
root = buildTree(inOrder, preOrder, 0, len(inOrder)-1)
 
# Let us test the build tree by printing Inorder traversal
print "Inorder traversal of the constructed tree is"
printInorder(root)
 
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)


C#




// C# program to construct a tree using
// inorder and preorder traversal
using System;
 
/* A binary tree node has data, pointer
to left child and a pointer to right child */
public class Node {
    public char data;
    public Node left, right;
 
    public Node(char item)
    {
        data = item;
        left = right = null;
    }
}
 
class GFG {
    public Node root;
    public static int preIndex = 0;
 
    /* Recursive function to construct binary
of size len from Inorder traversal in[]
and Preorder traversal pre[]. Initial values
of inStrt and inEnd should be 0 and len -1.
The function doesn't do any error checking for
cases where inorder and preorder do not form a tree */
    public virtual Node buildTree(char[] arr, char[] pre,
                                  int inStrt, int inEnd)
    {
        if (inStrt > inEnd) {
            return null;
        }
 
        /* Pick current node from Preorder traversal
     using preIndex and increment preIndex */
        Node tNode = new Node(pre[preIndex++]);
 
        /* If this node has no children then return */
        if (inStrt == inEnd) {
            return tNode;
        }
 
        /* Else find the index of this
       node in Inorder traversal */
        int inIndex = search(arr, inStrt,
                             inEnd, tNode.data);
 
        /* Using index in Inorder traversal,
    construct left and right subtress */
        tNode.left = buildTree(arr, pre, inStrt, inIndex - 1);
        tNode.right = buildTree(arr, pre, inIndex + 1, inEnd);
 
        return tNode;
    }
 
    /* UTILITY FUNCTIONS */
 
    /* Function to find index of value in arr[start...end]
The function assumes that value is present in in[] */
    public virtual int search(char[] arr, int strt,
                              int end, char value)
    {
        int i;
        for (i = strt; i <= end; i++) {
            if (arr[i] == value) {
                return i;
            }
        }
        return i;
    }
 
    /* This function is here just to test buildTree() */
    public virtual void printInorder(Node node)
    {
        if (node == null) {
            return;
        }
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        Console.Write(node.data + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        GFG tree = new GFG();
        char[] arr = new char[] { 'D', 'B', 'E', 'A', 'F', 'C' };
        char[] pre = new char[] { 'A', 'B', 'D', 'E', 'C', 'F' };
        int len = arr.Length;
        Node root = tree.buildTree(arr, pre, 0, len - 1);
 
        // building the tree by printing inorder traversal
        Console.WriteLine("Inorder traversal of "
                          + "constructed tree is : ");
        tree.printInorder(root);
    }
}
 
// This code is contributed by Shrikant13


Javascript




<script>
 
// Javascript program to construct a
// tree using inorder and preorder traversal
  
// A binary tree node has data, pointer
// to left child and a pointer to right child
class Node
{
    constructor(item)
    {
        this.data = item;
        this.left = this.right = null;
    }
}
 
let root;
let preIndex = 0;
 
// Recursive function to construct binary
// of size len from Inorder traversal in[]
// and Preorder traversal pre[]. Initial
// values of inStrt and inEnd should be 0
// and len -1. The function doesn't do any
// error checking for cases where inorder
// and preorder do not form a tree
function buildTree(In, pre, inStrt, inEnd)
{
    if (inStrt > inEnd)
        return null;
 
    // Pick current node from Preorder
    // traversal using preIndex and
    // increment preIndex
    let tNode = new Node(pre[preIndex++]);
 
    // If this node has no children then return
    if (inStrt == inEnd)
        return tNode;
 
    // Else find the index of this
    // node in Inorder traversal
    let inIndex = search(In, inStrt,
                         inEnd, tNode.data);
 
    // Using index in Inorder traversal,
    // construct left and right subtress
    tNode.left = buildTree(In, pre, inStrt,
                           inIndex - 1);
    tNode.right = buildTree(In, pre,
                            inIndex + 1,
                            inEnd);
 
    return tNode;
}
 
/* UTILITY FUNCTIONS */
  
// Function to find index of value
// in arr[start...end]. The function
// assumes that value is present in in[]
function search(arr, strt, end, value)
{
    let i;
    for(i = strt; i <= end; i++)
    {
        if (arr[i] == value)
            return i;
    }
    return i;
}
 
// This function is here just
// to test buildTree()
function printInorder(node)
{
    if (node == null)
            return;
  
    // First recur on left child
    printInorder(node.left);
 
    // Then print the data of node
    document.write(node.data + " ");
 
    // Now recur on right child
    printInorder(node.right);
}
 
// Driver code
let In = [ 'D', 'B', 'E', 'A', 'F', 'C' ];
let pre = [ 'A', 'B', 'D', 'E', 'C', 'F'];
let len = In.length;
root = buildTree(In, pre, 0, len - 1);
 
// Building the tree by printing
// inorder traversal
document.write("Inorder traversal of " +
               "constructed tree is : <br>");
printInorder(root);
 
// This code is contributed by patel2127
 
</script>


Output: 

Inorder traversal of the constructed tree is 
D B E A F C

 

Time Complexity: O(n^2). The worst case occurs when the tree is left-skewed. Example Preorder and Inorder traversals for worst-case are {A, B, C, D} and {D, C, B, A}. 
 

 

Efficient Approach : 
We can optimize the above solution using hashing (unordered_map in C++ or HashMap in Java). We store indexes of inorder traversal in a hash table. So that search can be done O(1) time. 

C++




/* C++ program to construct tree using inorder
   and preorder traversals */
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node {
    char data;
    struct Node* left;
    struct Node* right;
};
 
struct Node* newNode(char data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
 
/* Recursive function to construct binary of size
len from Inorder traversal in[] and Preorder traversal
pre[]. Initial values of inStrt and inEnd should be
0 and len -1. The function doesn't do any error
checking for cases where inorder and preorder
do not form a tree */
struct Node* buildTree(char in[], char pre[], int inStrt,
                       int inEnd, unordered_map<char, int>& mp)
{
    static int preIndex = 0;
 
    if (inStrt > inEnd)
        return NULL;
 
    /* Pick current node from Preorder traversal using preIndex
    and increment preIndex */
    char curr = pre[preIndex++];
    struct Node* tNode = newNode(curr);
 
    /* If this node has no children then return */
    if (inStrt == inEnd)
        return tNode;
 
    /* Else find the index of this node in Inorder traversal */
    int inIndex = mp[curr];
 
    /* Using index in Inorder traversal, construct left and
    right subtress */
    tNode->left = buildTree(in, pre, inStrt, inIndex - 1, mp);
    tNode->right = buildTree(in, pre, inIndex + 1, inEnd, mp);
 
    return tNode;
}
 
// This function mainly creates an unordered_map, then
// calls buildTree()
struct Node* buldTreeWrap(char in[], char pre[], int len)
{
    // Store indexes of all items so that we
    // we can quickly find later
    unordered_map<char, int> mp;
    for (int i = 0; i < len; i++)
        mp[in[i]] = i;
 
    return buildTree(in, pre, 0, len - 1, mp);
}
 
/* This function is here just to test buildTree() */
void printInorder(struct Node* node)
{
    if (node == NULL)
        return;
    printInorder(node->left);
    printf("%c ", node->data);
    printInorder(node->right);
}
 
/* Driver program to test above functions */
int main()
{
    char in[] = { 'D', 'B', 'E', 'A', 'F', 'C' };
    char pre[] = { 'A', 'B', 'D', 'E', 'C', 'F' };
    int len = sizeof(in) / sizeof(in[0]);
 
    struct Node* root = buldTreeWrap(in, pre, len);
 
    /* Let us test the built tree by printing
      Inorder traversal */
    printf("Inorder traversal of the constructed tree is \n");
    printInorder(root);
}


Java




/* Java program to construct tree using inorder
   and preorder traversals */
import java.io.*;
import java.util.*;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
class Node
{
  char data;
  Node left,right;
  Node(char item)
  {
    data = item;
    left = right = null;
  }
}
class Tree
{
 
  public static Node root;
 
  // Store indexes of all items so that we
  // we can quickly find later
  static HashMap<Character,Integer> mp = new HashMap<>();
  static int preIndex = 0;
 
  /* Recursive function to construct binary of size
    len from Inorder traversal in[] and Preorder traversal
    pre[]. Initial values of inStrt and inEnd should be
    0 and len -1. The function doesn't do any error
    checking for cases where inorder and preorder
    do not form a tree */
  public static Node buildTree(char[] in, char[] pre, int inStrt, int inEnd)
  {
 
    if(inStrt > inEnd)
    {
      return null;
    }
 
    /* Pick current node from Preorder traversal using preIndex
        and increment preIndex */
    char curr = pre[preIndex++];
    Node tNode;
    tNode = new Node(curr);
 
    /* If this node has no children then return */
    if (inStrt == inEnd)
    {
      return tNode;
    }
 
    /* Else find the index of this node in Inorder traversal */
    int inIndex = mp.get(curr);
 
    /* Using index in Inorder traversal, construct left and
        right subtress */
    tNode.left = buildTree(in, pre, inStrt, inIndex - 1);
    tNode.right = buildTree(in, pre, inIndex + 1, inEnd);
    return tNode;
  }
 
  // This function mainly creates an unordered_map, then
  // calls buildTree()
  public static Node buldTreeWrap(char[] in, char[] pre, int len)
  {
    for(int i = 0; i < len; i++)
    {
      mp.put(in[i], i);
    }
    return buildTree(in, pre, 0, len - 1);
  }
 
  /* This function is here just to test buildTree() */
  static void printInorder(Node node)
  {
    if(node == null)
    {
      return;
    }
    printInorder(node.left);
    System.out.print(node.data + " ");
    printInorder(node.right);
  }
 
  /* Driver code */
  public static void main (String[] args)
  {
    char[] in = {'D', 'B', 'E', 'A', 'F', 'C'};
    char[] pre = {'A', 'B', 'D', 'E', 'C', 'F'};
    int len = in.length;
 
    Tree.root=buldTreeWrap(in, pre, len);
 
    /* Let us test the built tree by printing
        Inorder traversal */
    System.out.println("Inorder traversal of the constructed tree is");
    printInorder(root);
  }
}
 
// This code is contributed by avanitrachhadiya2155


Python3




# Python3 program to construct tree using inorder
# and preorder traversals
 
# A binary tree node has data, pointer to left child
# and a pointer to right child
class Node:
     
    def __init__(self, x):
         
        self.data = x
        self.left = None
        self.right = None
 
# Recursive function to construct binary of size
# len from Inorder traversal in[] and Preorder traversal
# pre[]. Initial values of inStrt and inEnd should be
# 0 and len -1. The function doesn't do any error
# checking for cases where inorder and preorder
# do not form a tree
def buildTree(inn, pre, inStrt, inEnd):
     
    global preIndex, mp
 
    if (inStrt > inEnd):
        return None
 
    # Pick current node from Preorder traversal
    # using preIndex and increment preIndex
    curr = pre[preIndex]
    preIndex += 1
    tNode = Node(curr)
 
    # If this node has no children then return
    if (inStrt == inEnd):
        return tNode
 
    # Else find the index of this
    # node in Inorder traversal
    inIndex = mp[curr]
 
    # Using index in Inorder traversal,
    # construct left and right subtress
    tNode.left = buildTree(inn, pre, inStrt,
                           inIndex - 1)
    tNode.right = buildTree(inn, pre, inIndex + 1,
                            inEnd)
 
    return tNode
 
# This function mainly creates an
# unordered_map, then calls buildTree()
def buldTreeWrap(inn, pre, lenn):
     
    global mp
     
    # Store indexes of all items so that we
    # we can quickly find later
    # unordered_map<char, int> mp;
    for i in range(lenn):
        mp[inn[i]] = i
 
    return buildTree(inn, pre, 0, lenn - 1)
 
# This function is here just to test buildTree()
def prInorder(node):
 
    if (node == None):
        return
         
    prInorder(node.left)
    print(node.data, end = " ")
    prInorder(node.right)
 
# Driver code
if __name__ == '__main__':
     
    mp = {}
    preIndex = 0
 
    inn = [ 'D', 'B', 'E', 'A', 'F', 'C' ]
    pre = [ 'A', 'B', 'D', 'E', 'C', 'F' ]
    lenn = len(inn)
 
    root = buldTreeWrap(inn, pre,lenn)
 
    # Let us test the built tree by printing
    # Inorder traversal
    print("Inorder traversal of "
          "the constructed tree is")
     
    prInorder(root)
 
# This code is contributed by mohit kumar 29


C#




/* C# program to construct tree using inorder
   and preorder traversals */
using System;
using System.Collections.Generic;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
public class Node
{
    public char data;
    public Node left, right;
   
    public Node(char d)
    {
        data = d;
        left = right = null;
    }
}
public class Tree
{
    public static Node root;
   
    // Store indexes of all items so that we
    // we can quickly find later
    static Dictionary<char,int> mp = new Dictionary<char,int>();
    static int preIndex = 0;
   
    /* Recursive function to construct binary of size
    len from Inorder traversal in[] and Preorder traversal
    pre[]. Initial values of inStrt and inEnd should be
    0 and len -1. The function doesn't do any error
    checking for cases where inorder and preorder
    do not form a tree */
    static Node buildTree(char[] In, char[] pre,
                          int inStrt, int inEnd)
    {
        if(inStrt > inEnd)
        {
            return null;
        }
       
        /* Pick current node from Preorder traversal using preIndex
        and increment preIndex */
        char curr = pre[preIndex++];
        Node tNode;
        tNode = new Node(curr);
       
        /* If this node has no children then return */
        if(inStrt == inEnd)
        {
            return tNode;
        }
       
        /* Else find the index of this node in Inorder traversal */
        int inIndex = mp[curr];
       
        /* Using index in Inorder traversal, construct left and
        right subtress */
        tNode.left = buildTree(In, pre, inStrt, inIndex - 1);
        tNode.right = buildTree(In, pre, inIndex + 1, inEnd);
        return tNode;
    }
   
    // This function mainly creates an unordered_map, then
    // calls buildTree()
    public static Node buldTreeWrap(char[] In, char[] pre, int len)
    {
        for(int i = 0; i < len; i++)
        {
            mp.Add(In[i], i);
        }
        return buildTree(In, pre, 0, len - 1);
    }
   
    /* This function is here just to test buildTree() */
    static void printInorder(Node node)
    {
        if(node == null)
        {
            return;
        }
        printInorder(node.left);
        Console.Write(node.data + " ");
        printInorder(node.right);
    }
   
    /* Driver code */
    static public void Main (){
        char[] In = {'D', 'B', 'E', 'A', 'F', 'C'};
        char[] pre = {'A', 'B', 'D', 'E', 'C', 'F'};
        int len = In.Length;
        Tree.root = buldTreeWrap(In, pre, len);
       
        /* Let us test the built tree by printing
        Inorder traversal */
        Console.WriteLine("Inorder traversal of the constructed tree is");
        printInorder(Tree.root);
    }
}
 
// This code is contributed by rag2127


Output: 

Inorder traversal of the constructed tree is 
D B E A F C

 

Time Complexity : O(n)
 
Another approach : 
Use the fact that InOrder traversal is Left-Root-Right and PreOrder traversal is Root-Left-Right. Also, the first node in the PreOrder traversal is always the root node and the first node in the InOrder traversal is the leftmost node in the tree.
Maintain two data structures: Stack (to store the path we visited while traversing PreOrder array) and Set (to maintain the node in which the next right subtree is expected).

1. Do below until you reach the leftmost node. 
Keep creating the nodes from the PreOrder traversal 
If the stack’s topmost element is not in the set, link the created node to the left child of the stack’s topmost element (if any), without popping the element. 
Else link the created node to the right child of the stack’s topmost element. Remove the stack’s topmost element from the set and the stack. 
Push the node to a stack.
 

1

2. Keep popping the nodes from the stack until either the stack is empty, or the topmost element of the stack compares to the current element of InOrder traversal. Once the loop is over, push the last node back into the stack and into the set.
 

2

3. Goto Step 1.
 

3

 

C++




// C++ program to construct a tree using
// inorder and preorder traversal
#include<bits/stdc++.h>
using namespace std;
 
class TreeNode
{
    public:
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) { val = x; }
};
 
set<TreeNode*> s;
stack<TreeNode*> st;
 
// Function to build tree using given traversal
TreeNode* buildTree(int preorder[], int inorder[],int n)
{
 
    TreeNode* root = NULL;
    for (int pre = 0, in = 0; pre < n;)
    {
 
        TreeNode* node = NULL;
        do
        {
            node = new TreeNode(preorder[pre]);
            if (root == NULL)
            {
                root = node;
            }
            if (st.size() > 0)
            {
                if (s.find(st.top()) != s.end())
                {
                    s.erase(st.top());
                    st.top()->right = node;
                    st.pop();
                }
                else
                {
                    st.top()->left = node;
                }
            }
            st.push(node);
        } while (preorder[pre++] != inorder[in] && pre < n);
 
        node = NULL;
        while (st.size() > 0 && in < n &&
                st.top()->val == inorder[in])
        {
            node = st.top();
            st.pop();
            in++;
        }
 
        if (node != NULL)
        {
            s.insert(node);
            st.push(node);
        }
    }
 
    return root;
}
 
// Function to print tree in Inorder
void printInorder(TreeNode* node)
{
    if (node == NULL)
        return;
 
    /* first recur on left child */
    printInorder(node->left);
 
    /* then print the data of node */
    cout << node->val << " ";
 
    /* now recur on right child */
    printInorder(node->right);
}
 
// Driver code
int main()
{
    int in[] = { 9, 8, 4, 2, 10, 5, 10, 1, 6, 3, 13, 12, 7 };
    int pre[] = { 1, 2, 4, 8, 9, 5, 10, 10, 3, 6, 7, 12, 13 };
    int len = sizeof(in)/sizeof(int);
 
    TreeNode *root = buildTree(pre, in, len);
 
    printInorder(root);
    return 0;
}
 
// This code is contributed by Arnab Kundu


Java




// Java program to construct a tree using inorder and preorder traversal
import java.util.*;
 
public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}
 
class BinaryTree {
    static Set<TreeNode> set = new HashSet<>();
    static Stack<TreeNode> stack = new Stack<>();
 
    // Function to build tree using given traversal
    public TreeNode buildTree(int[] preorder, int[] inorder)
    {
 
        TreeNode root = null;
        for (int pre = 0, in = 0; pre < preorder.length;) {
 
            TreeNode node = null;
            do {
                node = new TreeNode(preorder[pre]);
                if (root == null) {
                    root = node;
                }
                if (!stack.isEmpty()) {
                    if (set.contains(stack.peek())) {
                        set.remove(stack.peek());
                        stack.pop().right = node;
                    }
                    else {
                        stack.peek().left = node;
                    }
                }
                stack.push(node);
            } while (preorder[pre++] != inorder[in] && pre < preorder.length);
 
            node = null;
            while (!stack.isEmpty() && in < inorder.length &&
                    stack.peek().val == inorder[in]) {
                node = stack.pop();
                in++;
            }
 
            if (node != null) {
                set.add(node);
                stack.push(node);
            }
        }
 
        return root;
    }
 
    // Function to print tree in Inorder
    void printInorder(TreeNode node)
    {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        System.out.print(node.val + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    // driver program to test above functions
    public static void main(String args[])
    {
        BinaryTree tree = new BinaryTree();
 
        int in[] = new int[] { 9, 8, 4, 2, 10, 5, 10, 1, 6, 3, 13, 12, 7 };
        int pre[] = new int[] { 1, 2, 4, 8, 9, 5, 10, 10, 3, 6, 7, 12, 13 };
        int len = in.length;
 
        TreeNode root = tree.buildTree(pre, in);
 
        tree.printInorder(root);
    }
}


Python3




# Python3 program to construct a tree using
# inorder and preorder traversal
class TreeNode:
     
    def __init__(self, x):     
        self.val = x
        self.left = None
        self.right = None
 
s = set()
st = []
 
# Function to build tree using given traversal
def buildTree(preorder, inorder, n):
    root = None;
     
    pre = 0
    in_t = 0
     
    while pre < n:
        node = None;
         
        while True:   
            node = TreeNode(preorder[pre])
             
            if (root == None):           
                root = node;
              
            if (len(st) > 0):       
                if (st[-1] in s):               
                    s.discard(st[-1]);
                    st[-1].right = node;
                    st.pop();
             
                else:               
                    st[-1].left = node;
                         
            st.append(node);
             
            if pre>=n or preorder[pre] == inorder[in_t]:
                pre += 1
                break
            pre += 1
             
        node = None;
         
        while (len(st) > 0 and in_t < n and st[-1].val == inorder[in_t]):       
            node = st[-1];
            st.pop();
            in_t += 1
     
 
        if (node != None):       
            s.add(node);
            st.append(node);
 
    return root;
 
# Function to print tree in_t Inorder
def printInorder( node):
 
    if (node == None):
        return;
 
    ''' first recur on left child '''
    printInorder(node.left);
 
    ''' then prthe data of node '''
    print(node.val, end=" ");
 
    ''' now recur on right child '''
    printInorder(node.right);
 
# Driver code
if __name__=='__main__':
     
    in_t = [ 9, 8, 4, 2, 10, 5, 10, 1, 6, 3, 13, 12, 7 ]
    pre = [ 1, 2, 4, 8, 9, 5, 10, 10, 3, 6, 7, 12, 13 ]
    l = len(in_t)
 
    root = buildTree(pre, in_t, l);
 
    printInorder(root);
     
    # This code is contributed by rutvik_56.


C#




// C# program to construct a tree
// using inorder and preorder traversal
using System;
using System.Collections.Generic;
 
public class TreeNode
{
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int x) { val = x; }
}
 
class GFG
{
    static HashSet<TreeNode> set = new HashSet<TreeNode>();
    static Stack<TreeNode> stack = new Stack<TreeNode>();
 
    // Function to build tree using given traversal
    public TreeNode buildTree(int[] preorder, int[] inorder)
    {
        TreeNode root = null;
        for (int pre = 0, iN = 0; pre < preorder.Length;)
        {
            TreeNode node = null;
            do {
                node = new TreeNode(preorder[pre]);
                if (root == null)
                {
                    root = node;
                }
                if (stack.Count != 0)
                {
                    if (set.Contains(stack.Peek()))
                    {
                        set.Remove(stack.Peek());
                        stack.Pop().right = node;
                    }
                    else
                    {
                        stack.Peek().left = node;
                    }
                }
                stack.Push(node);
            } while (preorder[pre++] != inorder[iN] &&
                     pre < preorder.Length);
 
            node = null;
            while (stack.Count != 0 && iN < inorder.Length &&
                   stack.Peek().val == inorder[iN])
            {
                node = stack.Pop();
                iN++;
            }
            if (node != null)
            {
                set.Add(node);
                stack.Push(node);
            }
        }
        return root;
    }
 
    // Function to print tree in Inorder
    void printInorder(TreeNode node)
    {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        Console.Write(node.val + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    // Driver Code
    public static void Main(String []args)
    {
        GFG tree = new GFG();
 
        int []iN = new int[] { 9, 8, 4, 2, 10, 5, 10,
                                 1, 6, 3, 13, 12, 7 };
        int []pre = new int[] { 1, 2, 4, 8, 9, 5, 10,
                                 10, 3, 6, 7, 12, 13 };
        int len = iN.Length;
 
        TreeNode root = tree.buildTree(pre, iN);
 
        tree.printInorder(root);
    }
}
 
// This code is contributed by 29AjayKumar


Javascript




<script>
// Javascript program to construct a tree using inorder and preorder traversal
     
      
    class TreeNode
    {
        constructor(x) {
            this.val = x;
            this.left = null;
            this.right = null;
          }
    }
     
    let set = new Set();
    let stack = [];
     
    // Function to build tree using given traversal
    function buildTree(preorder,inorder)
    {
        let root = null;
         
        for (let pre = 0,In=0; pre < preorder.length;) {
  
            let node = null;
            do {
                node = new TreeNode(preorder[pre]);
                if (root == null) {
                    root = node;
                }
                if (stack.length!=0) {
                    if (set.has(stack[stack.length-1])) {
                        set.delete(stack[stack.length-1]);
                        stack.pop().right = node;
                    }
                    else {
                        stack[stack.length-1].left = node;
                    }
                }
                stack.push(node);
            } while (preorder[pre++] != inorder[In] && pre < preorder.length);
  
            node = null;
            while (stack.length!=0 && In < inorder.length &&
                    stack[stack.length-1].val == inorder[In]) {
                node = stack.pop();
                In++;
            }
  
            if (node != null) {
                set.add(node);
                stack.push(node);
            }
        }
  
        return root;
    }
     
    // Function to print tree in Inorder
    function printInorder(node)
    {
        if (node == null)
            return;
  
        /* first recur on left child */
        printInorder(node.left);
  
        /* then print the data of node */
        document.write(node.val + " ");
  
        /* now recur on right child */
        printInorder(node.right);
    }
     
    // driver program to test above functions
    let In=[9, 8, 4, 2, 10, 5, 10, 1, 6, 3, 13, 12, 7 ];
    let pre=[1, 2, 4, 8, 9, 5, 10, 10, 3, 6, 7, 12, 13 ];
    let len = In.length;
    let root=buildTree(pre, In);
    printInorder(root);
     
     
    // This code is contributed by unknown2108
</script>


Output: 

9 8 4 2 10 5 10 1 6 3 13 12 7

 

Construct a Binary Tree from Postorder and Inorder
 




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