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Height of binary tree considering even level leaves only

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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 = 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

                    

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 diagram
var 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).
 



Last Updated : 29 Jul, 2022
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