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Smallest Subtree with all the Deepest Nodes

Problem statement​

Given the root of a binary tree, the depth of each node is the shortest distance to the root.

Return the smallest subtree such that it contains all the deepest nodes in the original tree.

A node is called the deepest if it has the largest depth possible among any node in the entire tree.

The subtree of a node is a tree consisting of that node, plus the set of all descendants of that node.

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4]Output: [2,7,4]Explanation: We return the node with value 2, colored in yellow in the diagram.The nodes coloured in blue are the deepest nodes of the tree.Notice that nodes 5, 3 and 2 contain the deepest nodes in the tree but node 2 is the smallest subtree among them, so we return it.

Example 2:

Input: root = [1]Output: [1]Explanation: The root is the deepest node in the tree.

Example 3:

Input: root = [0,1,3,null,2]Output: [2]Explanation: The deepest node in the tree is 2, the valid subtrees are the subtrees of nodes 2, 1 and 0 but the subtree of node 2 is the smallest.

Constraints:

  • The number of nodes in the tree will be in the range [1, 500].
  • 0 <= Node.val <= 500
  • The values of the nodes in the tree are unique.

Note: This question is the same as 1123: https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/

My solution​

/**
 * 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) {
    let deepestNode = null;
    let maxLevel = 0;
    
    function dfs(node, level) {
        if (!node) {
            return level;
        }
        
        const leftMaxLevel = dfs(node.left, level + 1)
        const rightMaxLevel = dfs(node.right, level + 1)
        const currentMaxLevel = Math.max(leftMaxLevel, rightMaxLevel);
        
        if (maxLevel < currentMaxLevel) {
            maxLevel = currentMaxLevel;
        }
        
        if (leftMaxLevel === rightMaxLevel && maxLevel === currentMaxLevel) {
            deepestNode = node;
            maxLevel = currentMaxLevel;
        }
        
        return currentMaxLevel;
    }
    
    dfs(root, 0)
    
    return deepestNode;
};