Deepest left leaf node in a binary tree | iterative approach
Given a Binary Tree, find the deepest leaf node that is left child of its parent. For example, consider the following tree. The deepest left leaf node is the node with value 9.
Examples:
Input :
1
/ \
2 3
/ / \
4 5 6
\ \
7 8
/ \
9 10
Output : 9
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. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.
Recursive approach to this problem is discussed here
For iterative approach, idea is similar to Method 2 of level order traversal
The idea is to traverse the tree iteratively and whenever a left tree node is pushed to queue, check if it is leaf node, if it’s leaf node, then update the result. Since we go level by level, the last stored leaf node is deepest one,
C++
// CPP program to find deepest left leaf// node of binary tree#include <bits/stdc++.h>using namespace std;// tree nodestruct Node { int data; Node *left, *right;};// returns a new tree NodeNode* newNode(int data){ Node* temp = new Node(); temp->data = data; temp->left = temp->right = NULL; return temp;}// return the deepest left leaf node// of binary treeNode* getDeepestLeftLeafNode(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(); // Since we go level by level, the last // stored left leaf node is deepest one, if (temp->left) { q.push(temp->left); if (!temp->left->left && !temp->left->right) result = temp->left; } if (temp->right) q.push(temp->right); } return result;}// driver programint 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); Node* result = getDeepestLeftLeafNode(root); if (result) cout << "Deepest Left Leaf Node :: " << result->data << endl; else cout << "No result, left leaf not found\n"; return 0;} |
Java
// Java program to find deepest left leaf// node of binary treeimport java.util.*;class GFG{// tree nodestatic class Node{ int data; Node left, right;};// returns a new tree Nodestatic Node newNode(int data){ Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp;}// return the deepest left leaf node// of binary treestatic Node getDeepestLeftLeafNode(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.remove(); // Since we go level by level, the last // stored left leaf node is deepest one, if (temp.left != null) { q.add(temp.left); if (temp.left.left == null && temp.left.right == null) result = temp.left; } if (temp.right != null) q.add(temp.right); } return result;}// Driver Codepublic 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); Node result = getDeepestLeftLeafNode(root); if (result != null) System.out.println("Deepest Left Leaf Node :: " + result.data); else System.out.println("No result, " + "left leaf not found"); }}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to find deepest# left leaf 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 deepest# left leaf nodedef getDeepestLeftLeafNode(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) if (not temp.left.left and not temp.left.right): result = temp.left # Since we go level by level, # the last stored right leaf # node is deepest one if (temp.right): q.append(temp.right) return result# Driver Codeif __name__ == '__main__': # create a binary tree 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) result = getDeepestLeftLeafNode(root) if result: print("Deepest Left Leaf Node ::", result.data) else: print("No result, Left leaf not found") # This code is contributed by# Shubham Singh(SHUBHAMSINGH10) |
C#
// C# program to find deepest left leaf// node of binary treeusing System;using System.Collections.Generic; class GFG{// tree nodeclass Node{ public int data; public Node left, right;};// returns a new tree Nodestatic Node newNode(int data){ Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp;}// return the deepest left leaf node// of binary treestatic Node getDeepestLeftLeafNode(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(); // Since we go level by level, the last // stored left leaf node is deepest one, if (temp.left != null) { q.Enqueue(temp.left); if (temp.left.left == null && temp.left.right == null) result = temp.left; } if (temp.right != null) q.Enqueue(temp.right); } return result;}// Driver Codepublic 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); Node result = getDeepestLeftLeafNode(root); if (result != null) Console.WriteLine("Deepest Left Leaf Node :: " + result.data); else Console.WriteLine("No result, " + "left leaf not found"); }}// This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript program to find deepest // left leaf node of binary tree class Node { constructor(data) { this.left = null; this.right = null; this.data = data; } } // returns a new tree Node function newNode(data) { let temp = new Node(data); return temp; } // return the deepest left leaf node // of binary tree function getDeepestLeftLeafNode(root) { if (root == null) return null; // create a queue for level order traversal let q = []; q.push(root); let result = null; // traverse until the queue is empty while (q.length > 0) { let temp = q[0]; q.shift(); // Since we go level by level, the last // stored left leaf node is deepest one, if (temp.left != null) { q.push(temp.left); if (temp.left.left == null && temp.left.right == null) result = temp.left; } if (temp.right != null) q.push(temp.right); } return result; } // construct a tree let 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); let result = getDeepestLeftLeafNode(root); if (result != null) document.write("Deepest Left Leaf Node :: " + result.data); else document.write("No result, " + "left leaf not found");</script> |
Output:
Deepest Left Leaf Node :: 9
