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
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,
Algorithm:
- Define a Node structure that has data, left, and right pointers.
- Create a function newNode that takes data as an argument and returns a new node with data set to data, and left and right pointers set to NULL.
- Create a function getDeepestLeftLeafNode that takes the root of a binary tree as an argument and returns the deepest left leaf node of the tree.
- Initialize a queue q and add the root node of the binary tree to the queue.
- Initialize a pointer result to NULL.
- Use a while loop to traverse the binary tree level by level until the queue becomes empty.
- Within the while loop, pop the front element of the queue and store it in a temporary pointer temp.
- If temp has a left child, add it to the queue, and check if it is a leaf node (i.e., it does not have left or right child).
- If the left child of temp is a leaf node, set result to temp->left.
- If temp has a right child, add it to the queue.
- Return result which will contain the deepest left leaf node of the binary tree.
- In the main function, construct a binary tree and call getDeepestLeftLeafNode to find the deepest left leaf node of the tree.
- If result is not NULL, print the data of the deepest left leaf node, otherwise print “No result, left leaf not found”.
Implementation:
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.*;public 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 pointers. 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 Code if __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
Time complexity: O(n) where n is no of nodes in a given binary tree
Auxiliary space: O(n)
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