Construct BST from given preorder traversal | Set 2

• Difficulty Level : Hard
• Last Updated : 06 Jul, 2021

Given preorder traversal of a binary search tree, construct the BST.
For example, if the given traversal is {10, 5, 1, 7, 40, 50}, then the output should be root of following tree.

10
/   \
5     40
/  \      \
1    7      50

We have discussed O(n^2) and O(n) recursive solutions in the previous post. Following is a stack based iterative solution that works in O(n) time.
1. Create an empty stack.
2. Make the first value as root. Push it to the stack.
3. Keep on popping while the stack is not empty and the next value is greater than stack’s top value. Make this value as the right child of the last popped node. Push the new node to the stack.
4. If the next value is less than the stack’s top value, make this value as the left child of the stack’s top node. Push the new node to the stack.
5. Repeat steps 2 and 3 until there are items remaining in pre[].

C++

 // A O(n) iterative program for construction of BST from preorder traversal #include using namespace std;   /* A binary tree node has data, pointer to left child and a pointer to right child */ class Node {     public:     int data;     Node *left, *right; } node;   // A Stack has array of Nodes, capacity, and top class Stack {     public:     int top;     int capacity;     Node** array; } stack;   // A utility function to create a new tree node Node* newNode( int data ) {     Node* temp = new Node();     temp->data = data;     temp->left = temp->right = NULL;     return temp; }   // A utility function to create a stack of given capacity Stack* createStack( int capacity ) {     Stack* stack = new Stack();     stack->top = -1;     stack->capacity = capacity;     stack->array = new Node*[stack->capacity * sizeof( Node* )];     return stack; }   // A utility function to check if stack is full int isFull( Stack* stack ) {     return stack->top == stack->capacity - 1; }   // A utility function to check if stack is empty int isEmpty( Stack* stack ) {     return stack->top == -1; }   // A utility function to push an item to stack void push( Stack* stack, Node* item ) {     if( isFull( stack ) )         return;     stack->array[ ++stack->top ] = item; }   // A utility function to remove an item from stack Node* pop( Stack* stack ) {     if( isEmpty( stack ) )         return NULL;     return stack->array[ stack->top-- ]; }   // A utility function to get top node of stack Node* peek( Stack* stack ) {     return stack->array[ stack->top ]; }   // The main function that constructs BST from pre[] Node* constructTree ( int pre[], int size ) {     // Create a stack of capacity equal to size     Stack* stack = createStack( size );       // The first element of pre[] is always root     Node* root = newNode( pre );       // Push root     push( stack, root );       int i;     Node* temp;       // Iterate through rest of the size-1 items of given preorder array     for ( i = 1; i < size; ++i )     {         temp = NULL;           /* Keep on popping while the next value is greater than         stack's top value. */         while ( !isEmpty( stack ) && pre[i] > peek( stack )->data )             temp = pop( stack );           // Make this greater value as the right child                 // and push it to the stack         if ( temp != NULL)         {             temp->right = newNode( pre[i] );             push( stack, temp->right );         }           // If the next value is less than the stack's top                 // value, make this value as the left child of the                 // stack's top node. Push the new node to stack         else         {             peek( stack )->left = newNode( pre[i] );             push( stack, peek( stack )->left );         }     }       return root; }     // A utility function to print inorder traversal of a Binary Tree void printInorder (Node* node) {     if (node == NULL)         return;     printInorder(node->left);     cout<data<<" ";     printInorder(node->right); }   // Driver program to test above functions int main () {     int pre[] = {10, 5, 1, 7, 40, 50};     int size = sizeof( pre ) / sizeof( pre );       Node *root = constructTree(pre, size);       cout<<"Inorder traversal of the constructed tree: \n";     printInorder(root);       return 0; }   //This code is contributed by rathbhupendra

C

 // A O(n) iterative program for construction of BST from preorder traversal #include #include #include   /* A binary tree node has data, pointer to left child    and a pointer to right child */ typedef struct Node {     int data;     struct Node *left, *right; } Node;   // A Stack has array of Nodes, capacity, and top typedef struct Stack {     int top;     int capacity;     Node* *array; } Stack;   // A utility function to create a new tree node Node* newNode( int data ) {     Node* temp = (Node *)malloc( sizeof( Node ) );     temp->data = data;     temp->left = temp->right = NULL;     return temp; }   // A utility function to create a stack of given capacity Stack* createStack( int capacity ) {     Stack* stack = (Stack *)malloc( sizeof( Stack ) );     stack->top = -1;     stack->capacity = capacity;     stack->array = (Node **)malloc( stack->capacity * sizeof( Node* ) );     return stack; }   // A utility function to check if stack is full int isFull( Stack* stack ) {     return stack->top == stack->capacity - 1; }   // A utility function to check if stack is empty int isEmpty( Stack* stack ) {     return stack->top == -1; }   // A utility function to push an item to stack void push( Stack* stack, Node* item ) {     if( isFull( stack ) )         return;     stack->array[ ++stack->top ] = item; }   // A utility function to remove an item from stack Node* pop( Stack* stack ) {     if( isEmpty( stack ) )         return NULL;     return stack->array[ stack->top-- ]; }   // A utility function to get top node of stack Node* peek( Stack* stack ) {     return stack->array[ stack->top ]; }   // The main function that constructs BST from pre[] Node* constructTree ( int pre[], int size ) {     // Create a stack of capacity equal to size     Stack* stack = createStack( size );       // The first element of pre[] is always root     Node* root = newNode( pre );       // Push root     push( stack, root );       int i;     Node* temp;       // Iterate through rest of the size-1 items of given preorder array     for ( i = 1; i < size; ++i )     {         temp = NULL;           /* Keep on popping while the next value is greater than            stack's top value. */         while ( !isEmpty( stack ) && pre[i] > peek( stack )->data )             temp = pop( stack );           // Make this greater value as the right child         // and push it to the stack         if ( temp != NULL)         {             temp->right = newNode( pre[i] );             push( stack, temp->right );         }           // If the next value is less than the stack's top         // value, make this value as the left child of the         // stack's top node. Push the new node to stack         else         {             peek( stack )->left = newNode( pre[i] );             push( stack, peek( stack )->left );         }     }       return root; }     // A utility function to print inorder traversal of a Binary Tree void printInorder (Node* node) {     if (node == NULL)         return;     printInorder(node->left);     printf("%d ", node->data);     printInorder(node->right); }   // Driver program to test above functions int main () {     int pre[] = {10, 5, 1, 7, 40, 50};     int size = sizeof( pre ) / sizeof( pre );       Node *root = constructTree(pre, size);       printf("Inorder traversal of the constructed tree: \n");     printInorder(root);       return 0; }

Java

 // Java program to construct BST from given preorder traversal   import java.util.*;   // A binary tree node class Node {       int data;     Node left, right;       Node(int d) {         data = d;         left = right = null;     } }   class BinaryTree {       // The main function that constructs BST from pre[]     Node constructTree(int pre[], int size) {           // The first element of pre[] is always root         Node root = new Node(pre);           Stack s = new Stack();           // Push root         s.push(root);           // Iterate through rest of the size-1 items of given preorder array         for (int i = 1; i < size; ++i) {             Node temp = null;               /* Keep on popping while the next value is greater than              stack's top value. */             while (!s.isEmpty() && pre[i] > s.peek().data) {                 temp = s.pop();             }               // Make this greater value as the right child             // and push it to the stack             if (temp != null) {                 temp.right = new Node(pre[i]);                 s.push(temp.right);             }                           // If the next value is less than the stack's top             // value, make this value as the left child of the             // stack's top node. Push the new node to stack             else {                 temp = s.peek();                 temp.left = new Node(pre[i]);                 s.push(temp.left);             }         }           return root;     }       // A utility function to print inorder traversal of a Binary Tree     void printInorder(Node node) {         if (node == null) {             return;         }         printInorder(node.left);         System.out.print(node.data + " ");         printInorder(node.right);     }       // Driver program to test above functions     public static void main(String[] args) {         BinaryTree tree = new BinaryTree();         int pre[] = new int[]{10, 5, 1, 7, 40, 50};         int size = pre.length;         Node root = tree.constructTree(pre, size);         System.out.println("Inorder traversal of the constructed tree is ");         tree.printInorder(root);     } }   // This code has been contributed by Mayank Jaiswal

Python3

 # Python3 program to construct BST # from given preorder traversal   # A binary tree node class Node:       def __init__(self, data = 0):         self.data = data         self.left = None         self.right = None   class BinaryTree :       # The main function that constructs BST from pre[]     def constructTree(self, pre, size):           # The first element of pre[] is always root         root = Node(pre)           s = []           # append root         s.append(root)           i = 1           # Iterate through rest of the size-1         # items of given preorder array         while ( i < size):             temp = None               # Keep on popping while the next value             # is greater than stack's top value.             while (len(s) > 0 and pre[i] > s[-1].data):                 temp = s.pop()                           # Make this greater value as the right child             # and append it to the stack             if (temp != None):                 temp.right = Node(pre[i])                 s.append(temp.right)                           # If the next value is less than the stack's top             # value, make this value as the left child of the             # stack's top node. append the new node to stack             else :                 temp = s[-1]                 temp.left = Node(pre[i])                 s.append(temp.left)             i = i + 1                   return root           # A utility function to print     # inorder traversal of a Binary Tree     def printInorder(self,node):         if (node == None):             return                   self.printInorder(node.left)         print(node.data, end = " ")         self.printInorder(node.right)   # Driver code tree = BinaryTree() pre = [10, 5, 1, 7, 40, 50] size = len(pre) root = tree.constructTree(pre, size) print("Inorder traversal of the constructed tree is ") tree.printInorder(root)   # This code is contributed by Arnab Kundu

C#

 using System; using System.Collections.Generic;   // c# program to construct BST from given preorder traversal   // A binary tree node public class Node {       public int data;     public Node left, right;       public Node(int d)     {         data = d;         left = right = null;     } }   public class BinaryTree {       // The main function that constructs BST from pre[]     public virtual Node constructTree(int[] pre, int size)     {           // The first element of pre[] is always root         Node root = new Node(pre);           Stack s = new Stack();           // Push root         s.Push(root);           // Iterate through rest of the size-1 items of given preorder array         for (int i = 1; i < size; ++i)         {             Node temp = null;               /* Keep on popping while the next value is greater than              stack's top value. */             while (s.Count > 0 && pre[i] > s.Peek().data)             {                 temp = s.Pop();             }               // Make this greater value as the right child             // and push it to the stack             if (temp != null)             {                 temp.right = new Node(pre[i]);                 s.Push(temp.right);             }               // If the next value is less than the stack's top             // value, make this value as the left child of the             // stack's top node. Push the new node to stack             else             {                 temp = s.Peek();                 temp.left = new Node(pre[i]);                 s.Push(temp.left);             }         }           return root;     }       // A utility function to print inorder traversal of a Binary Tree     public virtual void printInorder(Node node)     {         if (node == null)         {             return;         }         printInorder(node.left);         Console.Write(node.data + " ");         printInorder(node.right);     }       // Driver program to test above functions     public static void Main(string[] args)     {         BinaryTree tree = new BinaryTree();         int[] pre = new int[]{10, 5, 1, 7, 40, 50};         int size = pre.Length;         Node root = tree.constructTree(pre, size);         Console.WriteLine("Inorder traversal of the constructed tree is ");         tree.printInorder(root);     } }     // This code is contributed by Shrikant13

Javascript



Output:

Inorder traversal of the constructed tree is
1 5 7 10 40 50

Time Complexity: O(n). The complexity looks more from first look. If we take a closer look, we can observe that every item is pushed and popped only once. So at most 2n push/pop operations are performed in the main loops of constructTree(). Therefore, time complexity is O(n).