[Leetcode] 333. Largest BST Subtree 解题报告

题目：

Given a binary tree, find the largest subtree which is a Binary Search Tree (BST), where largest means subtree with largest number of nodes in it.

Note:
A subtree must include all of its descendants.
Here's an example:

    10    / \   5  15  / \   \  1   8   7

The Largest BST Subtree in this case is the highlighted one. 
The return value is the subtree's size, which is 3.

Follow up:
Can you figure out ways to solve it with O(n) time complexity?

思路：

一道bottom-up的深度优先搜索题目，也就是首先判断左右子树分别是不是合法的BST。如果是的话，我们还需要哪些信息呢？首先是需要知道左右子树的范围，因为我们要判断当前结点为根的树是否为二叉搜索树，就要满足当前结点大于左子树的最大值，并且小于右子树的最小值。其次我们还需要分别知道其左右子树上的最大合法BST的大小。有了这些信息，就可以判断以当前结点为根的二叉树是否为二叉搜索树了。

实现过程中有个小技巧，就是让DFS返回vector<int>。我原来传入了三个int的引用，写的代码又臭又长。。。

代码：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int largestBSTSubtree(TreeNode* root) {        int ans = 0;        DFS(root, ans);        return ans;    }private:    vector<int> DFS(TreeNode *root, int &ans) {        if (!root) {            return vector<int>{0, INT_MAX, INT_MIN};        }        vector<int> left = DFS(root->left, ans);        vector<int> right = DFS(root->right, ans);        if (root->val > left[2] && root->val < right[1]) {            int min_value = min(root->val, left[1]);            int max_value = max(root->val, right[2]);            ans = max(ans, left[0] + right[0] + 1);            return vector<int>{left[0] + right[0] + 1, min_value, max_value};        }        return vector<int> {0, INT_MIN, INT_MAX};    }};