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; else 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; }

Output:

Height of tree is 4

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

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