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
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; } |
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Java
// Java program to find deepest right 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 the deepest right leaf node // of binary tree static Node getDeepestRightLeafNode(Node root) { if (root == null) return null; // create a queue for level order traversal Queue<Node> q = new LinkedList<>(); q.add(root); Node result = null; // traverse until the queue is empty while (!q.isEmpty()) { Node temp = q.peek(); q.poll(); if (temp.left != null) { q.add(temp.left); } // Since we go level by level, the last // stored right leaf node is deepest one if (temp.right != null) { q.add(temp.right); if (temp.right.left == null && temp.right.right == null) result = temp.right; } } 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.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 != null) System.out.println("Deepest Right Leaf Node :: " + result.data); else System.out.println("No result, right leaf not found\n"); } } /* This code is contributed by PrinciRaj1992 */ |
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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) |
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C#
// C# program to find deepest right 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 the deepest right leaf node // of binary tree static Node getDeepestRightLeafNode(Node root) { if (root == null) return null; // Create a queue for level order traversal Queue<Node> q = new Queue<Node>(); q.Enqueue(root); Node result = null; // Traverse until the queue is empty while (q.Count!=0) { Node temp = q.Peek(); q.Dequeue(); if (temp.left != null) { q.Enqueue(temp.left); } // Since we go level by level, the last // stored right leaf node is deepest one if (temp.right != null) { q.Enqueue(temp.right); if (temp.right.left == null && temp.right.right == null) result = temp.right; } } 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.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 != null) Console.WriteLine("Deepest Right Leaf Node :: " + result.data); else Console.WriteLine("No result, right leaf not found\n"); } } // This code is contributed by Princi Singh |
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Output:
Deepest Right Leaf Node :: 10
Time Complexity : O(n)
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