题目：
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.
思路：
题意大概是找最深叶子节点的最低公共祖先，所以我们要存的有两个，一个是节点，另一个是对应的深度。对于每个节点，如果为空直接返回；否则判别左右子的深度，如果左和右相等，返回当前节点，否则谁深度大返回谁。因为要记录两个，所以递归式返回一个pair较为方便。
代码：
/**
* 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 depth(root).second;
}
pair<int, TreeNode*> depth(TreeNode* root)
{
if(!root)
return {0,root};
auto left = depth(root->left);
auto right =depth(root->right);
int ld=left.first, rd=right.first;
if(ld>rd)
return {ld+1,left.second};
if(ld<rd)
return {rd+1,right.second};
return {ld+1,root};
}
};