Height of binary tree considering even level leaves only

Find the height of the binary tree given that only the nodes on the even levels are considered as the valid leaf nodes.
The height of a binary tree is the number of edges between the tree’s root and its furthest leaf. But what if we bring a twist and change the definition of a leaf node. Let us define a valid leaf node as the node that has no children and is at an even level (considering root node as an odd level node).

Output :Height of tree is 4

Solution : The approach to this problem is slightly different from the normal height finding approach. In the return step, we check if the node is a valid root node or not. If it is valid, return 1, else we return 0. Now in the recursive step- if the left and the right sub-tree both yield 0, the current node yields 0 too, because in that case there is no path from current node to a valid leaf node. But in case at least one of the values returned by the children is non-zero, it means the leaf node on that path is a valid leaf node, and hence that path can contribute to the final result, so we return max of the values returned + 1 for the current node.

/* Program to find height of the tree considering
   only even level leaves. */
#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 {
    int data;
    struct Node* left;
    struct Node* right;

int heightOfTreeUtil(Node* root, bool isEven)
    // Base Case
    if (!root)
        return 0;

    if (!root->left && !root->right) {
        if (isEven)
            return 1;
            return 0;

    /*left stores the result of left subtree,
      and right stores the result of right subtree*/
    int left = heightOfTreeUtil(root->left, !isEven);
    int right = heightOfTreeUtil(root->right, !isEven);

    /*If both left and right returns 0, it means
      there is no valid path till leaf node*/
    if (left == 0 && right == 0)
        return 0;

    return (1 + max(left, right));

/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct Node* newNode(int data)
    struct Node* node = 
              (struct Node*)malloc(sizeof(struct Node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;

    return (node);

int heightOfTree(Node* root)
    return heightOfTreeUtil(root, false);

/* Driver program to test above functions*/
int main()
    // Let us create binary tree shown in above diagram
    struct Node* root = newNode(1);

    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->left->right->left = newNode(6);
    cout << "Height of tree is " << heightOfTree(root);
    return 0;


Height of tree is 4

Time Complexity:O(n) where n is number of nodes in given binary tree.

My Personal Notes arrow_drop_up

In love with a semicolon because sometimes i miss it so badly)

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Practice Tags :
Article Tags :
Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:

3.3 Average Difficulty : 3.3/5.0
Based on 6 vote(s)

User Actions