Find the height of the binary tree given that only the nodes on the even levels are considered 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 the root node as an odd level node).
Output: Height of the tree is 4
Approach: 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 the current node to a valid leaf node. But in the 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 the max of the values returned + 1 for the current node.
C++
/* 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;} |
Java
/* Java Program to find height of the tree considering only even level leaves. */class GfG {/* A binary tree node has data, pointer to left child and a pointer to right child */static class Node { int data; Node left; Node right; }static int heightOfTreeUtil(Node root, boolean isEven) { // Base Case if (root == null) return 0; if (root.left == null && root.right == null) { if (isEven == true) 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 + Math.max(left, right)); } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return (node); } static int heightOfTree(Node root) { return heightOfTreeUtil(root, false); } /* Driver program to test above functions*/public static void main(String[] args) { // Let us create binary tree shown in above diagram 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); System.out.println("Height of tree is " + heightOfTree(root)); }} |
Python3
# Program to find height of the tree considering # only even level leaves.# Helper class that allocates a new node with the # given data and None left and right pointers. class newNode: def __init__(self, data): self.data = data self.left = None self.right = Nonedef heightOfTreeUtil(root, isEven): # Base Case if (not root): return 0 if (not root.left and not root.right): if (isEven): return 1 else: return 0 # left stores the result of left subtree, # and right stores the result of right subtree left = heightOfTreeUtil(root.left, not isEven) right = heightOfTreeUtil(root.right, not isEven) #If both left and right returns 0, it means # there is no valid path till leaf node if (left == 0 and right == 0): return 0 return (1 + max(left, right))def heightOfTree(root): return heightOfTreeUtil(root, False)# Driver Codeif __name__ == '__main__': # Let us create binary tree shown # in above diagram 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) print("Height of tree is", heightOfTree(root))# This code is contributed by PranchalK |
C#
/* C# Program to find height of the tree considering only even level leaves. */using System;class GfG{ /* A binary tree node has data, pointer to left child and a pointer to right child */ class Node { public int data; public Node left; public Node right; } static int heightOfTreeUtil(Node root, bool isEven) { // Base Case if (root == null) return 0; if (root.left == null && root.right == null) { if (isEven == true) 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 + Math.Max(left, right)); } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return (node); } static int heightOfTree(Node root) { return heightOfTreeUtil(root, false); } /* Driver code*/ public static void Main(String[] args) { // Let us create binary tree // shown in above diagram 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); Console.WriteLine("Height of tree is " + heightOfTree(root)); } } /* This code is contributed by Rajput-Ji*/ |
Javascript
<script>/* javascript Program to find height of the tree considering only even level leaves. */ /* * A binary tree node has data, pointer to left child and a pointer to right * child */ class Node { constructor(val) { this.data = val; this.left = null; this.right = null; } } function heightOfTreeUtil(root, isEven) { // Base Case if (root == null) return 0; if (root.left == null && root.right == null) { if (isEven == true) return 1; else return 0; } /* * left stores the result of left subtree, and right stores the result of right * subtree */ var left = heightOfTreeUtil(root.left, !isEven); var 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 + Math.max(left, right)); } /* * Helper function that allocates a new node with the given data and NULL left * and right pointers. */ function newNode(data) {var node = new Node(); node.data = data; node.left = null; node.right = null; return (node); } function heightOfTree(root) { return heightOfTreeUtil(root, false); } /* Driver program to test above functions */ // Let us create binary tree shown in above diagramvar 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); document.write("Height of tree is " + heightOfTree(root));// This code is contributed by umadevi9616 </script> |
Output
Height of tree is 4
Time Complexity: O(n), Where n is the number of nodes in a given binary tree.
Auxiliary Space: O(n), If we don’t consider the size of the recursive call stack for function calls then O(1), otherwise O(n).
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