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.
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; } |
chevron_right
filter_none
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)); } } |
chevron_right
filter_none
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 = None def 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 Code if __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 |
chevron_right
filter_none
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*/ |
chevron_right
filter_none
Output:
Height of tree is 4
Time Complexity:O(n) where n is number of nodes in given binary tree.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
