LeetCode解题报告（242）-- 865. Smallest Subtree with all the Deepest Nodes

Problem

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:

1
2
3

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

Example 3:

1
2
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.

Analysis

这是一道二叉树高度相关的题目，要求找到包含最深节点的最小子树。遇到这类题目很自然地就会往递归方面考虑。利用递归计算树的高度并不难，这道题目难的地方在于如何找到最深的那棵树，是否需要把节点记录下来？

我们先从题目给的example入手分析一下，可以看到对于以2为根的这颗树来说，2的左子树和右子树的深度都是1。往2的父节点看，以5为根的这棵树来说，它的左子树深度（2）比右子树（3）要小，所以选择的是它右子树。实际上题目给出的这个例子非常好，它把两种情况都覆盖了。对于某个根节点root来说：

如果它的左、右子树高度相等：那么它就是要返回的节点，因为两个子树包含的节点的深度都是一样的；
如果它的左、右子树高度不相等，那么它就不可能是最小的子树的根节点。所有就要往深度更大的那个子树去寻找。

Solution

写一个递归的函数去计算深度，再写一个递归去计算比较每个节点的左右子树，相当于是两个递归依次进行。

Code

/**
 * 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) {
        int rootDiff = getDepth(root->left) - getDepth(root->right);
        return rootDiff == 0 ? root : (rootDiff > 0 ? subtreeWithAllDeepest(root->left) : subtreeWithAllDeepest(root->right));
    }
private:
    int getDepth(TreeNode* root) {
        return !root ? 0 : max(getDepth(root->left), getDepth(root->right)) + 1;
    }
};

Summary

这道题目虽然看上去要求有点复杂，但是本质还是二叉树深度相关的内容，利用递归能够很方便解决。这道题目的分享到这里，谢谢您的支持！