Lowest common ancestor of deepest leaves

https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves

Given a rooted binary tree, return the lowest common ancestor of its deepest leaves.

Recall that:

The node of a binary tree is a leaf if and only if it has no children The depth of the root of the tree is 0, and if the depth of a node is d, the depth of each of its children is d+1. The lowest common ancestor of a set S of nodes is the node A with the largest depth such that every node in S is in the subtree with root A.

Example 1:

Input: root = [1,2,3]
Output: [1,2,3]

Explanation: 
The deepest leaves are the nodes with values 2 and 3.
The lowest common ancestor of these leaves is the node with value 1.
The answer returned is a TreeNode object (not an array) with serialization "[1,2,3]".

Example 2:

Input: root = [1,2,3,4]
Output: [4]

Example 3:

Input: root = [1,2,3,4,5]
Output: [2,4,5]

Constraints:

The given tree will have between 1 and 1000 nodes.
Each node of the tree will have a distinct value between 1 and 1000.

Examples : Lowest Common Ancestor node of deepest leaves is highlighted in orange color.

e.g. 1

e.g. 2

e.g. 3

e.g. 4

e.g. 5

Approach : Two pass through binary tree

First Pass : In first pass we move from root to deepest leaves by Level order traversal using BFS. After first pass, deepestLevelNodes queue will have leaf nodes at the deepest level
Second Pass : In second pass we move upwards in the tree, from nodes at the deepest level to root node. To move upwards, we add the parent of the node in deepestLevelNodes queue. We keep doing this untill there is only one TreeNode in deepestLevelNodes queue.

Implementation : Iterative

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode lcaDeepestLeaves(TreeNode root) {
        if(root == null)
            return root;
        
        Map<TreeNode, TreeNode> parentMap = new HashMap<>();
        parentMap.put(root, null);
        Queue<TreeNode> q = new ArrayDeque<>();
        q.add(root);
        Queue<TreeNode> deepestLevelNodes = null;
        // First Pass : Level order traversal using BFS
        // After first pass, deepestLevelNodes queue will have leaf nodes at the deepest level
        while(!q.isEmpty()) {
            deepestLevelNodes = new ArrayDeque<>(q); 
            int size = q.size();
            for(int i = 0; i < size; i++) {
                TreeNode node = q.poll();
                if(node.left != null) {
                   parentMap.put(node.left, node); 
                   q.add(node.left);   
                }
                if(node.right != null) {
                    parentMap.put(node.right, node);
                    q.add(node.right);
                }
            }
        }
        
        // Second Pass : We move upwards in the tree, from nodes at the deepest level to root node.
        // To move upwards, we add the parent of the node in deepestLevelNodes queue.
        // We keep doing this untill there is only one TreeNode in deepestLevelNodes queue.
        while(deepestLevelNodes.size() > 1) {
            Set<TreeNode> set = new HashSet<>();
            int size = deepestLevelNodes.size();
            for(int i = 0; i < size; i++) {
                set.add(parentMap.get(deepestLevelNodes.poll()));
            }
            for(TreeNode node : set) 
                deepestLevelNodes.add(node);
        }
        
        return deepestLevelNodes.poll();
        
    }
}

References :

https://www.youtube.com/watch?v=QF7ZBH8mXHE

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