Calculate height of Binary Tree using Inorder and Level Order Traversal

Given inorder traversal and Level Order traversal of a Binary Tree. The task is to calculate the height of the tree without constructing it. 

Example:

Input : Input: Two arrays that represent Inorder
       and level order traversals of a 
       Binary Tree
       in[]    = {4, 8, 10, 12, 14, 20, 22};
       level[] = {20, 8, 22, 4, 12, 10, 14};
Output : 4

The binary tree that can be constructed from the 
given traversals is:

We can clearly see in the above image that the 
height of the tree is 4.

The approach to calculating height is similar to the approach discussed in the post Constructing Tree from Inorder and Level Order Traversals.
Let us consider the above example.
in[] = {4, 8, 10, 12, 14, 20, 22}; 
level[] = {20, 8, 22, 4, 12, 10, 14};
In a Levelorder sequence, the first element is the root of the tree. So we know ’20’ is root for given sequences. By searching ’20’ in Inorder sequence, we can find out all elements on the left side of ‘20’ are in left subtree and elements on right are in the right subtree. So we know below structure now.  

             20
           /    \
          /      \ 
 {4, 8, 10, 12, 14}  {22}   

Let us call {4, 8, 10, 12, 14} as left subarray in Inorder traversal and {22} as right subarray in Inorder traversal. 
In level order traversal, keys of left and right subtrees are not consecutive. So we extract all nodes from level order traversal which are in left subarray of Inorder traversal. To calculate the height of the left subtree of the root, we recur for the extracted elements from level order traversal and left subarray of inorder traversal. In the above example, we recur for the following two arrays. 

// Recur for following arrays to 
// calculate the height of the left subtree
In[]    = {4, 8, 10, 12, 14}
level[] = {8, 4, 12, 10, 14} 

Similarly, we recur for the following two arrays and calculate the height of the right subtree.

// Recur for following arrays to calculate
// height of the right subtree
In[]    = {22}
level[] = {22} 

Below is the implementation of the above approach:  

C++

// C++ program to caulate height of Binary Tree
// from InOrder and LevelOrder Traversals
#include <iostream>
using namespace std;

/* Function to find index of value
   in the InOrder Traversal array */
int search(int arr[], int strt, int end, int value)
{
    for (int i = strt; i <= end; i++)
        if (arr[i] == value)
            return i;
    return -1;
}

// Function to calculate the height
// of the Binary Tree
int getHeight(int in[], int level[], int start,
              int end, int& height, int n)
{

    // Base Case
    if (start > end)
        return 0;

    // Get index of current root in InOrder Traversal
    int getIndex = search(in, start, end, level[0]);

    if (getIndex == -1)
        return 0;

    // Count elements in Left Subtree
    int leftCount = getIndex - start;

    // Count elements in right Subtree
    int rightCount = end - getIndex;

    // Declare two arrays for left and
    // right subtrees
    int* newLeftLevel = new int[leftCount];
    int* newRightLevel = new int[rightCount];

    int lheight = 0, rheight = 0;
    int k = 0;

    // Extract values from level order traversal array
    // for current left subtree
    for (int i = 0; i < n; i++) {
        for (int j = start; j < getIndex; j++) {
            if (level[i] == in[j]) {
                newLeftLevel[k] = level[i];
                k++;
                break;
            }
        }
    }

    k = 0;

    // Extract values from level order traversal array
    // for current right subtree
    for (int i = 0; i < n; i++) {
        for (int j = getIndex + 1; j <= end; j++) {
            if (level[i] == in[j]) {
                newRightLevel[k] = level[i];
                k++;
                break;
            }
        }
    }

    // Recursively call to calculate height of left Subtree
    if (leftCount > 0)
        lheight = getHeight(in, newLeftLevel, start,
                            getIndex - 1, height, leftCount);

    // Recursively call to calculate height of right Subtree
    if (rightCount > 0)
        rheight = getHeight(in, newRightLevel,
                            getIndex + 1, end, height, rightCount);

    // Current height
    height = max(lheight + 1, rheight + 1);

    // Delete Auxiliary arrays
    delete[] newRightLevel;
    delete[] newLeftLevel;

    // return height
    return height;
}

// Driver program to test above functions
int main()
{
    int in[] = { 4, 8, 10, 12, 14, 20, 22 };
    int level[] = { 20, 8, 22, 4, 12, 10, 14 };
    int n = sizeof(in) / sizeof(in[0]);

    int h = 0;

    cout << getHeight(in, level, 0, n - 1, h, n);

    return 0;
}

Java

// Java program to caulate height of Binary Tree
// from InOrder and LevelOrder Traversals
import java.util.*;
class GFG
{
static int height;

/* Function to find index of value
in the InOrder Traversal array */
static int search(int arr[], int strt,
                  int end, int value)
{
    for (int i = strt; i <= end; i++)
        if (arr[i] == value)
            return i;
    return -1;
}

// Function to calculate the height
// of the Binary Tree
static int getHeight(int in[], int level[],
                     int start, int end, int n)
{

    // Base Case
    if (start > end)
        return 0;

    // Get index of current root in InOrder Traversal
    int getIndex = search(in, start, end, level[0]);

    if (getIndex == -1)
        return 0;

    // Count elements in Left Subtree
    int leftCount = getIndex - start;

    // Count elements in right Subtree
    int rightCount = end - getIndex;

    // Declare two arrays for left and
    // right subtrees
    int []newLeftLevel = new int[leftCount];
    int []newRightLevel = new int[rightCount];

    int lheight = 0, rheight = 0;
    int k = 0;

    // Extract values from level order traversal array
    // for current left subtree
    for (int i = 0; i < n; i++)
    {
        for (int j = start; j < getIndex; j++)
        {
            if (level[i] == in[j])
            {
                newLeftLevel[k] = level[i];
                k++;
                break;
            }
        }
    }

    k = 0;

    // Extract values from level order traversal array
    // for current right subtree
    for (int i = 0; i < n; i++)
    {
        for (int j = getIndex + 1; j <= end; j++)
        {
            if (level[i] == in[j])
            {
                newRightLevel[k] = level[i];
                k++;
                break;
            }
        }
    }

    // Recursively call to calculate
    // height of left Subtree
    if (leftCount > 0)
        lheight = getHeight(in, newLeftLevel, start,
                  getIndex - 1, leftCount);

    // Recursively call to calculate
    // height of right Subtree
    if (rightCount > 0)
        rheight = getHeight(in, newRightLevel,
                  getIndex + 1, end, rightCount);

    // Current height
    height = Math.max(lheight + 1, rheight + 1);

    // Delete Auxiliary arrays
    newRightLevel=null;
    newLeftLevel=null;

    // return height
    return height;
}

// Driver program to test above functions
public static void main(String[] args)
{
    int in[] = {4, 8, 10, 12, 14, 20, 22};
    int level[] = {20, 8, 22, 4, 12, 10, 14};
    int n = in.length;

    height = 0;

    System.out.println(getHeight(in, level, 0, n - 1, n));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to calculate height of Binary Tree from
# InOrder and LevelOrder Traversals

'''
Function to find the index of value in the InOrder
Traversal list
'''

def search(arr, start, end, value):
    for i in range(start, end + 1):
        if arr[i] == value:
            return i
    return -1

'''
Function to calculate the height of the Binary Tree
'''
def getHeight(inOrder, levelOrder,
              start, end, height, n):

    # Base Case
    if start > end:
        return 0

    # Get Index of current root in InOrder Traversal
    getIndex = search(inOrder, start, end, levelOrder[0])

    if getIndex == -1:
        return 0

    # Count elements in Left Subtree
    leftCount = getIndex - start

    # Count elements in Right Subtree
    rightCount = end - getIndex

    # Declare two lists for left and right subtrees
    newLeftLevel = [None for _ in range(leftCount)]
    newRightLevel = [None for _ in range(rightCount)]

    lheight, rheight, k = 0, 0, 0

    # Extract values from level order traversal list
    # for current left subtree
    for i in range(n):
        for j in range(start, getIndex):
            if levelOrder[i] == inOrder[j]:
                newLeftLevel[k] = levelOrder[i]
                k += 1
                break

    k = 0

    # Extract values from level order traversal list
    # for current right subtree
    for i in range(n):
        for j in range(getIndex + 1, end + 1):
            if levelOrder[i] == inOrder[j]:
                newRightLevel[k] = levelOrder[i]
                k += 1
                break

    # Recursively call to calculate height
    # of left subtree
    if leftCount > 0:
        lheight = getHeight(inOrder, newLeftLevel, start,
                            getIndex - 1, height, leftCount)

    # Recursively call to calculate height
    # of right subtree
    if rightCount > 0:
        rheight = getHeight(inOrder, newRightLevel,
                            getIndex + 1, end, height, rightCount)

    # current height
    height = max(lheight + 1, rheight + 1)

    # return height
    return height

# Driver Code
if __name__=='__main__':
    inOrder = [4, 8, 10, 12, 14, 20, 22]
    levelOrder = [20, 8, 22, 4, 12, 10, 14]
    n, h = len(inOrder), 0
    print(getHeight(inOrder, levelOrder, 0, n - 1, h, n))

# This code is contributed by harshraj22

C#

// C# program to caulate height of Binary Tree
// from InOrder and LevelOrder Traversals
using System;
using System.Collections.Generic;

class GFG
{
static int height;

/* Function to find index of value
in the InOrder Traversal array */
static int search(int []arr, int strt,
                  int end, int value)
{
    for (int i = strt; i <= end; i++)
        if (arr[i] == value)
            return i;
    return -1;
}

// Function to calculate the height
// of the Binary Tree
static int getHeight(int []In, int []level,
                     int start, int end, int n)
{

    // Base Case
    if (start > end)
        return 0;

    // Get index of current root in
    // InOrder Traversal
    int getIndex = search(In, start, end, level[0]);

    if (getIndex == -1)
        return 0;

    // Count elements in Left Subtree
    int leftCount = getIndex - start;

    // Count elements in right Subtree
    int rightCount = end - getIndex;

    // Declare two arrays for left and
    // right subtrees
    int []newLeftLevel = new int[leftCount];
    int []newRightLevel = new int[rightCount];

    int lheight = 0, rheight = 0;
    int k = 0;

    // Extract values from level order traversal array
    // for current left subtree
    for (int i = 0; i < n; i++)
    {
        for (int j = start; j < getIndex; j++)
        {
            if (level[i] == In[j])
            {
                newLeftLevel[k] = level[i];
                k++;
                break;
            }
        }
    }

    k = 0;

    // Extract values from level order traversal array
    // for current right subtree
    for (int i = 0; i < n; i++)
    {
        for (int j = getIndex + 1; j <= end; j++)
        {
            if (level[i] == In[j])
            {
                newRightLevel[k] = level[i];
                k++;
                break;
            }
        }
    }

    // Recursively call to calculate
    // height of left Subtree
    if (leftCount > 0)
        lheight = getHeight(In, newLeftLevel, start,
                  getIndex - 1, leftCount);

    // Recursively call to calculate
    // height of right Subtree
    if (rightCount > 0)
        rheight = getHeight(In, newRightLevel,
                  getIndex + 1, end, rightCount);

    // Current height
    height = Math.Max(lheight + 1, rheight + 1);

    // Delete Auxiliary arrays
    newRightLevel = null;
    newLeftLevel = null;

    // return height
    return height;
}

// Driver Code
public static void Main(String[] args)
{
    int []In = {4, 8, 10, 12, 14, 20, 22};
    int []level = {20, 8, 22, 4, 12, 10, 14};
    int n = In.Length;

    height = 0;

    Console.WriteLine(getHeight(In, level, 0, n - 1, n));
}
}

// This code is contributed by Princi Singh

Javascript

<script>


//Javascript program to caulate height of Binary Tree
// from InOrder and LevelOrder Traversals

var height = 0;

/* Function to find index of value
in the InOrder Traversal array */
function search(arr, strt, end, value)
{
    for (var i = strt; i <= end; i++)
        if (arr[i] == value)
            return i;
    return -1;
}

// Function to calculate the height
// of the Binary Tree
function getHeight(In, level, start, end, n)
{

    // Base Case
    if (start > end)
        return 0;

    // Get index of current root in
    // InOrder Traversal
    var getIndex = search(In, start, end, level[0]);

    if (getIndex == -1)
        return 0;

    // Count elements in Left Subtree
    var leftCount = getIndex - start;

    // Count elements in right Subtree
    var rightCount = end - getIndex;

    // Declare two arrays for left and
    // right subtrees
    var newLeftLevel = Array(leftCount);
    var newRightLevel = Array(rightCount);

    var lheight = 0, rheight = 0;
    var k = 0;

    // Extract values from level order traversal array
    // for current left subtree
    for (var i = 0; i < n; i++)
    {
        for (var j = start; j < getIndex; j++)
        {
            if (level[i] == In[j])
            {
                newLeftLevel[k] = level[i];
                k++;
                break;
            }
        }
    }

    k = 0;

    // Extract values from level order traversal array
    // for current right subtree
    for (var i = 0; i < n; i++)
    {
        for (var j = getIndex + 1; j <= end; j++)
        {
            if (level[i] == In[j])
            {
                newRightLevel[k] = level[i];
                k++;
                break;
            }
        }
    }

    // Recursively call to calculate
    // height of left Subtree
    if (leftCount > 0)
        lheight = getHeight(In, newLeftLevel, start,
                  getIndex - 1, leftCount);

    // Recursively call to calculate
    // height of right Subtree
    if (rightCount > 0)
        rheight = getHeight(In, newRightLevel,
                  getIndex + 1, end, rightCount);

    // Current height
    height = Math.max(lheight + 1, rheight + 1);

    // Delete Auxiliary arrays
    newRightLevel = null;
    newLeftLevel = null;

    // return height
    return height;
}

// Driver Code
var In = [4, 8, 10, 12, 14, 20, 22];
var level = [20, 8, 22, 4, 12, 10, 14];
var n = In.length;
height = 0;
document.write(getHeight(In, level, 0, n - 1, n));


</script>

Output:  

4

Time Complexity: O(n^2)

In the worst case, the time complexity of the above algorithm will be O(n^2). The search() function takes O(n) time and getHeight() is called n times.

Space Complexity: O(n)

The space complexity of the above algorithm will be O(n). The auxiliary space used by the program is O(n) which is used to store the Level Order Traversal array.