Deepest right leaf node in a binary tree | Iterative approach

Given a Binary Tree, find the deepest leaf node that is right child of its parent. For example, consider the following tree. The deepest right leaf node is the node with value 10.

Examples:

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

Output : 10

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is similar to Method 2 of level order traversal

Traverse the tree level by level and while pushing right child to queue, check if it is leaf node, if it’s leaf node, then update the result and since we are traversing level by level, the last stored right leaf will be the deepest right leaf node.

C++

// CPP program to find deepest right 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 the deepest right leaf node 
// of binary tree 
Node* getDeepestRightLeafNode(Node* root) 
{ 
    if (!root) 
        return NULL; 
   
    // create a queue for level order traversal 
    queue<Node*> q; 
    q.push(root); 
   
    Node* result = NULL; 
   
    // traverse until the queue is empty 
    while (!q.empty()) { 
        Node* temp = q.front(); 
        q.pop(); 
   
          
        if (temp->left) { 
            q.push(temp->left); 
        } 
          
        // Since we go level by level, the last  
        // stored right leaf node is deepest one  
        if (temp->right){ 
            q.push(temp->right); 
            if (!temp->right->left && !temp->right->right) 
                result = temp->right; 
        } 
    } 
    return result; 
} 
   
// driver program 
int main() 
{ 
    // construct a tree 
    Node* root = newNode(1); 
    root->left = newNode(2); 
    root->right = newNode(3); 
    root->left->right = 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); 
   
    Node* result = getDeepestRightLeafNode(root); 
    if (result) 
        cout << "Deepest Right Leaf Node :: "
             << result->data << endl; 
    else
        cout << "No result, right leaf not found\n"; 
    return 0; 
} 

Python3

# Python3 program to find closest
# value in Binary search Tree

_MIN = -2147483648
_MAX = 2147483648

# Helper function that allocates a new
# node with the given data and None
# left and right poers.
class newnode:

# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None

# utility function to return level
# of given node
def getDeepestRightLeafNode(root) :

if (not root):
return None

# create a queue for level
# order traversal
q = []
q.append(root)

result = None

# traverse until the queue is empty
while (len(q)):
temp = q[0]
q.pop(0)

if (temp.left):
q.append(temp.left)

# Since we go level by level, the last
# stored right leaf node is deepest one
if (temp.right):
q.append(temp.right)
if (not temp.right.left and
not temp.right.right):
result = temp.right

return result

# Driver Code
if __name__ == ‘__main__’:

# create a binary tree
root = newnode(1)
root.left = newnode(2)
root.right = newnode(3)
root.left.right = 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)

result = getDeepestRightLeafNode(root)
if result:
print(“Deepest Right Leaf Node ::”,
result.data)
else:
print(“No result, right leaf not found”)

# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

Output: