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 tree consisting of that node, plus the set of all descendants of that node.

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

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.

Solution

Time complexity : O(n)
Space complexity : O(h)

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

    pair<int, TreeNode*> dfs(TreeNode* node) {
        if (!node) return {0, nullptr};
        pair<int, TreeNode*> l = dfs(node->left),
                            r = dfs(node->right);
        int d_l = l.first, d_r = r.first;
        return {max(l.first, r.first)+1, d_l == d_r ? node : d_l > d_r ? l.second : r.second};
    }
};

if node == nullptr，return {0, nullptr}

if left depth == right depth，左右皆有最深子樹，return {left.depth+1, node}。

if left depth > right depth，return {left.depth, left subtree}

if left depth < right depth，return {right.depth, right subtree}