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Find depth of the deepest odd level leaf node
  • Difficulty Level : Easy
  • Last Updated : 20 Nov, 2019

Write a code to get the depth of the deepest odd level leaf node in a binary tree. Consider that level starts with 1. Depth of a leaf node is number of nodes on the path from root to leaf (including both leaf and root).

For example, consider the following tree. The deepest odd level node is the node with value 9 and depth of this node is 5.

       1
     /   \
    2     3
  /      /  \  
 4      5    6
        \     \
         7     8
        /       \
       9         10
                 /
                11

The idea is to recursively traverse the given binary tree and while traversing, maintain a variable “level” which will store the current node’s level in the tree. If current node is leaf then check “level” is odd or not. If level is odd then return it. If current node is not leaf, then recursively find maximum depth in left and right subtrees, and return maximum of the two depths.

C++




// C++ program to find depth of the
// deepest odd level leaf node
#include <bits/stdc++.h>
using namespace std;
  
// A utility function to find 
// maximum of two integers
int max(int x, int y) 
{ 
    return (x > y)? x : y; 
}
  
// A Binary Tree node
struct Node
{
    int data;
    struct Node *left, *right;
};
  
// A utility function to allocate a
// new tree node
struct Node* newNode(int data)
{
    struct Node* node = (struct Node*) malloc(sizeof(struct Node));
    node->data = data;
    node->left = node->right = NULL;
    return node;
}
  
// A recursive function to find depth of
// the deepest odd level leaf
int depthOfOddLeafUtil(struct Node *root,
                               int level)
{
    // Base Case
    if (root == NULL)
        return 0;
  
    // If this node is a leaf and its level
    // is odd, return its level
    if (root->left == NULL && 
        root->right == NULL && level & 1)
        return level;
  
    // If not leaf, return the maximum value
    // from left and right subtrees
    return max(depthOfOddLeafUtil(root->left, level + 1),
               depthOfOddLeafUtil(root->right, level + 1));
}
  
/* Main function which calculates the depth
   of deepest odd level leaf. This function
   mainly uses depthOfOddLeafUtil() */
int depthOfOddLeaf(struct Node *root)
{
    int level = 1, depth = 0;
    return depthOfOddLeafUtil(root, level);
}
  
// Driver Code
int main()
{
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->right->left = newNode(5);
    root->right->right = newNode(6);
    root->right->left->right = newNode(7);
    root->right->right->right = newNode(8);
    root->right->left->right->left = newNode(9);
    root->right->right->right->right = newNode(10);
    root->right->right->right->right->left = newNode(11);
  
    cout << depthOfOddLeaf(root) 
         << " is the required depth";
    getchar();
    return 0;
}
  
// This code is contributed
// by Akanksha Rai
C



// C program to find depth of the deepest odd level leaf node
#include <stdio.h>
#include <stdlib.h>
  
// A utility function to find maximum of two integers
int max(int x, int y) { return (x > y)? x : y; }
  
// A Binary Tree node
struct Node
{
    int data;
    struct Node *left, *right;
};
  
// A utility function to allocate a new tree node
struct Node* newNode(int data)
{
    struct Node* node = (struct Node*) malloc(sizeof(struct Node));
    node->data = data;
    node->left = node->right = NULL;
    return node;
}
  
// A recursive function to find depth of the deepest odd level leaf
int depthOfOddLeafUtil(struct Node *root,int level)
{
    // Base Case
    if (root == NULL)
        return 0;
  
    // If this node is a leaf and its level is odd, return its level
    if (root->left==NULL && root->right==NULL && level&1)
        return level;
  
    // If not leaf, return the maximum value from left and right subtrees
    return max(depthOfOddLeafUtil(root->left, level+1),
            depthOfOddLeafUtil(root->right, level+1));
}
  
/* Main function which calculates the depth of deepest odd level leaf.
This function mainly uses depthOfOddLeafUtil() */
int depthOfOddLeaf(struct Node *root)
{
    int level = 1, depth = 0;
    return depthOfOddLeafUtil(root, level);
}
  
// Driver program to test above functions
int main()
{
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->right->left = newNode(5);
    root->right->right = newNode(6);
    root->right->left->right = newNode(7);
    root->right->right->right = newNode(8);
    root->right->left->right->left = newNode(9);
    root->right->right->right->right = newNode(10);
    root->right->right->right->right->left = newNode(11);
  
    printf("%d is the required depth\n", depthOfOddLeaf(root));
    getchar();
    return 0;
}
Java



// Java program to find depth of deepest odd level node
  
// A binary tree node
class Node 
{
    int data;
    Node left, right;
   
    Node(int item) 
    {
        data = item;
        left = right = null;
    }
}
   
class BinaryTree 
{
    Node root;
   
    // A recursive function to find depth of the deepest odd level leaf
    int depthOfOddLeafUtil(Node node, int level) 
    {
        // Base Case
        if (node == null)
            return 0;
   
        // If this node is a leaf and its level is odd, return its level
        if (node.left == null && node.right == null && (level & 1) != 0)
            return level;
   
        // If not leaf, return the maximum value from left and right subtrees
        return Math.max(depthOfOddLeafUtil(node.left, level + 1),
                depthOfOddLeafUtil(node.right, level + 1));
    }
   
    /* Main function which calculates the depth of deepest odd level leaf.
       This function mainly uses depthOfOddLeafUtil() */
    int depthOfOddLeaf(Node node) 
    {
        int level = 1, depth = 0;
        return depthOfOddLeafUtil(node, level);
    }
   
    public static void main(String args[]) 
    {
        int k = 45;
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.right.left = new Node(5);
        tree.root.right.right = new Node(6);
        tree.root.right.left.right = new Node(7);
        tree.root.right.right.right = new Node(8);
        tree.root.right.left.right.left = new Node(9);
        tree.root.right.right.right.right = new Node(10);
        tree.root.right.right.right.right.left = new Node(11);
        System.out.println(tree.depthOfOddLeaf(tree.root) + 
                                                   " is the required depth");
    }
}
   
// This code has been contributed by Mayank Jaiswal
Python



# Python program to find depth of the deepest odd level
# leaf node
  
# A Binary tree node
class Node:
      
    # Constructor to create a new node
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
  
# A recursive function to find depth of the deepest 
# odd level leaf node
def depthOfOddLeafUtil(root, level):
      
    # Base Case
    if root is None:
        return 0 
  
    # If this node is leaf and its level is odd, return
    # its level
    if root.left is None and root.right is None and level&1: 
        return level 
      
    # If not leaf, return the maximum value from left
    # and right subtrees
    return (max(depthOfOddLeafUtil(root.left, level+1),
                depthOfOddLeafUtil(root.right, level+1)))
  
# Main function which calculates the depth of deepest odd
# level leaf . 
# This function mainly uses depthOfOddLeafUtil()
def depthOfOddLeaf(root):
    level = 1 
    depth = 0
    return depthOfOddLeafUtil(root, level)
  
# Driver program to test above function
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.right.left = Node(5)
root.right.right = Node(6)
root.right.left.right = Node(7)
root.right.right.right = Node(8)
root.right.left.right.left = Node(9)
root.right.right.right.right = Node(10)
root.right.right.right.right.left= Node(11)
  
print "%d is the required depth" %(depthOfOddLeaf(root))
  
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
C#



using System;
  
// C# program to find depth of deepest odd level node 
  
// A binary tree node 
public class Node
{
    public int data;
    public Node left, right;
  
    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}
  
public class BinaryTree
{
    public Node root;
  
    // A recursive function to find depth of the deepest odd level leaf 
    public virtual int depthOfOddLeafUtil(Node node, int level)
    {
        // Base Case 
        if (node == null)
        {
            return 0;
        }
  
        // If this node is a leaf and its level is odd, return its level 
        if (node.left == null && node.right == null && (level & 1) != 0)
        {
            return level;
        }
  
        // If not leaf, return the maximum value from left and right subtrees 
        return Math.Max(depthOfOddLeafUtil(node.left, level + 1),
                        depthOfOddLeafUtil(node.right, level + 1));
    }
  
    /* Main function which calculates the depth of deepest odd level leaf. 
    This function mainly uses depthOfOddLeafUtil() */
    public virtual int depthOfOddLeaf(Node node)
    {
        int level = 1, depth = 0;
        return depthOfOddLeafUtil(node, level);
    }
  
    public static void Main(string[] args)
    {
        int k = 45;
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.right.left = new Node(5);
        tree.root.right.right = new Node(6);
        tree.root.right.left.right = new Node(7);
        tree.root.right.right.right = new Node(8);
        tree.root.right.left.right.left = new Node(9);
        tree.root.right.right.right.right = new Node(10);
        tree.root.right.right.right.right.left = new Node(11);
        Console.WriteLine(tree.depthOfOddLeaf(tree.root) + " is the required depth");
    }
}
  
// This code is contributed by Shrikant13

Output:
5 is the required depth
Time Complexity: The function does a simple traversal of the tree, so the complexity is O(n).


Iterative Approach
This approach is contributed by Mandeep Singh. Traverse the tree in iterative fashion for each level, and whenever you encounter the leaf node, check if level is odd, if level is odd, then update the result.
C++



// CPP program to find
// depth of the deepest
// odd level leaf node
// of binary tree
#include <bits/stdc++.h>
using namespace std;
  
// tree node
struct Node 
{
    int data;
    Node *left, *right;
};
  
// returns a new
// tree Node
Node* newNode(int data)
{
    Node* temp = new Node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
  
// return max odd number
// depth of leaf node
int maxOddLevelDepth(Node* root)
{
    if (!root)
        return 0;
  
    // create a queue
    // for level order 
    // traversal
    queue<Node*> q;
    q.push(root);
  
    int result = INT_MAX;
    int level = 0;
  
    // traverse until the
    // queue is empty
    while (!q.empty()) 
    {
        int size = q.size();
        level += 1;
  
        // traverse for
        // complete level
        while(size > 0)
        {
            Node* temp = q.front();
            q.pop();
  
            // check if the node is
            // leaf node and level 
            // is odd if level is
            // odd, then update result
            if(!temp->left && !temp->right 
                      && (level % 2 != 0))
            {
                result = level;
            }
          
            // check for left child
            if (temp->left) 
            {
                q.push(temp->left);
            }
              
            // check for right child
            if (temp->right)
            {
                q.push(temp->right);
            }
            size -= 1;
        } 
    }
      
    return result;
}
  
// driver program
int main()
{
    // construct a tree
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->right->left = newNode(5);
    root->right->right = newNode(6);
    root->right->left->right = newNode(7);
    root->right->right->right = newNode(8);
    root->right->left->right->left = newNode(9);
    root->right->right->right->right = newNode(10);
    root->right->right->right->right->left = newNode(11);
  
    int result = maxOddLevelDepth(root);
      
    if (result == INT_MAX)
        cout << "No leaf node at odd level\n";
    else
        cout << result;
        cout << " is the required depth " << endl;
    return 0;
}
Java



// Java program to find depth of the deepest
// odd level leaf node of binary tree
import java.util.*;
  
class GFG 
{
  
// tree node
static class Node 
{
    int data;
    Node left, right;
};
  
// returns a new tree Node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
  
// return max odd number depth of leaf node
static int maxOddLevelDepth(Node root)
{
    if (root == null)
        return 0;
  
    // create a queue for level order 
    // traversal
    Queue<Node> q = new LinkedList<>();
    q.add(root);
  
    int result = Integer.MAX_VALUE;
    int level = 0;
  
    // traverse until the queue is empty
    while (!q.isEmpty()) 
    {
        int size = q.size();
        level += 1;
  
        // traverse for complete level
        while(size > 0)
        {
            Node temp = q.peek();
            q.remove();
  
            // check if the node is leaf node and 
            // level is odd if level is odd,
            // then update result
            if(temp.left == null && 
               temp.right == null && (level % 2 != 0))
            {
                result = level;
            }
          
            // check for left child
            if (temp.left != null) 
            {
                q.add(temp.left);
            }
              
            // check for right child
            if (temp.right != null)
            {
                q.add(temp.right);
            }
            size -= 1;
        } 
    }
    return result;
}
  
// Driver Code
public static void main(String[] args)
{
    // construct a tree
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.right.left = newNode(5);
    root.right.right = newNode(6);
    root.right.left.right = newNode(7);
    root.right.right.right = newNode(8);
    root.right.left.right.left = newNode(9);
    root.right.right.right.right = newNode(10);
    root.right.right.right.right.left = newNode(11);
  
    int result = maxOddLevelDepth(root);
      
    if (result == Integer.MAX_VALUE)
        System.out.println("No leaf node at odd level");
    else
    {
        System.out.print(result);
        System.out.println(" is the required depth ");
    }
}
}
  
// This code is contributed by Rajput-Ji
Python3



# Python3 program to find depth of the deepest 
# odd level leaf node of binary tree
  
INT_MAX = 2**31
  
# tree node returns a new tree Node
class newNode: 
    def __init__(self, data): 
        self.data = data 
        self.left = self.right = None
          
# return max odd number depth 
# of leaf node 
def maxOddLevelDepth(root) :
  
    if (not root): 
        return 0
  
    # create a queue for level order 
    # traversal 
    q = [] 
    q.append(root) 
  
    result = INT_MAX 
    level = 0
  
    # traverse until the queue is empty 
    while (len(q)) :
      
        size = len(q) 
        level += 1
  
        # traverse for complete level 
        while(size > 0) :
          
            temp = q[0] 
            q.pop(0) 
  
            # check if the node is leaf node 
            # and level is odd if level is 
            # odd, then update result 
            if(not temp.left and not temp.right 
                    and (level % 2 != 0)) :
              
                result = level 
              
            # check for left child 
            if (temp.left) :
              
                q.append(temp.left) 
              
            # check for right child 
            if (temp.right) :
              
                q.append(temp.right) 
              
            size -= 1
      
    return result 
  
# Driver Code 
if __name__ == '__main__':
    root = newNode(1) 
    root.left = newNode(2) 
    root.right = newNode(3) 
    root.left.left = newNode(4) 
    root.right.left = newNode(5) 
    root.right.right = newNode(6) 
    root.right.left.right = newNode(7) 
    root.right.right.right = newNode(8) 
    root.right.left.right.left = newNode(9) 
    root.right.right.right.right = newNode(10) 
    root.right.right.right.right.left = newNode(11) 
  
    result = maxOddLevelDepth(root) 
    if (result == INT_MAX) :
        print("No leaf node at odd level") 
    else:
        print(result, end = "")
        print(" is the required depth ")
          
# This code is contributed
# by SHUBHAMSINGH10
C#



// C# program to find depth of the deepest
// odd level leaf node of binary tree
using System;
using System.Collections.Generic;
      
class GFG 
{
  
// tree node
public class Node 
{
    public int data;
    public Node left, right;
};
  
// returns a new tree Node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
  
// return max odd number depth of leaf node
static int maxOddLevelDepth(Node root)
{
    if (root == null)
        return 0;
  
    // create a queue for level order 
    // traversal
    Queue<Node> q = new Queue<Node>();
    q.Enqueue(root);
  
    int result = int.MaxValue;
    int level = 0;
  
    // traverse until the queue is empty
    while (q.Count != 0) 
    {
        int size = q.Count;
        level += 1;
  
        // traverse for complete level
        while(size > 0)
        {
            Node temp = q.Peek();
            q.Dequeue();
  
            // check if the node is leaf node and 
            // level is odd if level is odd,
            // then update result
            if(temp.left == null && 
               temp.right == null &&
              (level % 2 != 0))
            {
                result = level;
            }
          
            // check for left child
            if (temp.left != null) 
            {
                q.Enqueue(temp.left);
            }
              
            // check for right child
            if (temp.right != null)
            {
                q.Enqueue(temp.right);
            }
            size -= 1;
        } 
    }
    return result;
}
  
// Driver Code
public static void Main(String[] args)
{
    // construct a tree
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.right.left = newNode(5);
    root.right.right = newNode(6);
    root.right.left.right = newNode(7);
    root.right.right.right = newNode(8);
    root.right.left.right.left = newNode(9);
    root.right.right.right.right = newNode(10);
    root.right.right.right.right.left = newNode(11);
  
    int result = maxOddLevelDepth(root);
      
    if (result == int.MaxValue)
        Console.WriteLine("No leaf node at odd level");
    else
    {
        Console.Write(result);
        Console.WriteLine(" is the required depth ");
    }
}
}
  
// This code is contributed by PrinciRaj1992
Output:
5 is the required depth
Time Complexity: Time Complexity is O(n).
This article is contributed by Chandra Prakash. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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