865. Smallest Subtree with all the Deepest Nodes

/**
 * Problem: LeetCode 865 - Smallest Subtree with all the Deepest Nodes
 * Difficulty: Medium
 * Language: Java
 * Author: CodeByTushu
 */
/**
 * 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 {
    private static class Result {
        int depth;
        TreeNode node;

        Result(int depth, TreeNode node) {
            this.depth = depth;
            this.node = node;
        }
    }

    public TreeNode subtreeWithAllDeepest(TreeNode root) {
        return dfs(root).node;
    }

    private Result dfs(TreeNode node) {
        if (node == null) {
            return new Result(0, null);
        }

        Result left = dfs(node.left);
        Result right = dfs(node.right);

        if (left.depth == right.depth) {
            return new Result(left.depth + 1, node);
        } else if (left.depth > right.depth) {
            return new Result(left.depth + 1, left.node);
        } else {
            return new Result(right.depth + 1, right.node);
        }
    }
}

"""
Problem: LeetCode 865 - Smallest Subtree with all the Deepest Nodes
Difficulty: Medium
Language: Python
Author: CodeByTushu
"""
from typing import Optional, Tuple

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

class Solution:
    def subtreeWithAllDeepest(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        def dfs(node: Optional[TreeNode]) -> Tuple[int, Optional[TreeNode]]:
            if not node:
                return 0, None

            left_depth, left_node = dfs(node.left)
            right_depth, right_node = dfs(node.right)

            if left_depth == right_depth:
                return left_depth + 1, node
            elif left_depth > right_depth:
                return left_depth + 1, left_node
            else:
                return right_depth + 1, right_node

        return dfs(root)[1]

/**
 * Problem: LeetCode 865 - Smallest Subtree with all the Deepest Nodes
 * Difficulty: Medium
 * Language: C++
 * Author: CodeByTushu
 */
#include <utility>
#include <algorithm>

using namespace std;

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    pair<int, TreeNode*> dfs(TreeNode* node) {
        if (!node) {
            return {0, nullptr};
        }

        auto left = dfs(node->left);
        auto right = dfs(node->right);

        if (left.first == right.first) {
            return {left.first + 1, node};
        } else if (left.first > right.first) {
            return {left.first + 1, left.second};
        } else {
            return {right.first + 1, right.second};
        }
    }

public:
    TreeNode* subtreeWithAllDeepest(TreeNode* root) {
        return dfs(root).second;
    }
};

/**
 * Problem: LeetCode 865 - Smallest Subtree with all the Deepest Nodes
 * Difficulty: Medium
 * Language: JavaScript
 * Author: CodeByTushu
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var subtreeWithAllDeepest = function(root) {
    function dfs(node) {
        if (!node) {
            return { depth: 0, node: null };
        }

        const left = dfs(node.left);
        const right = dfs(node.right);

        if (left.depth === right.depth) {
            return { depth: left.depth + 1, node: node };
        } else if (left.depth > right.depth) {
            return { depth: left.depth + 1, node: left.node };
        } else {
            return { depth: right.depth + 1, node: right.node };
        }
    }

    return dfs(root).node;
};