【LeetCode】333. Largest BST Subtree 解题报告(C++)
- 作者: 负雪明烛
- id: fuxuemingzhu
- 个人博客:http://fuxuemingzhu.cn/
题目地址:https://leetcode-cn.com/problems/largest-bst-subtree/
题目描述
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.
Example:
Input: [10,5,15,1,8,null,7]
10
/ \
5 15
/ \ \
1 8 7
Output: 3
Explanation: 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?
题目大意
找出一个树中最大的BST有多少个节点。
解题方法
DFS
如果用简单的方法做,就是先判断每个树是不是BST,如果是则返回其节点个数,如果不是就递归他的左右子树,最后结果要用最大值。
C++代码如下:
/**
* 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) {
if (!root) return 0;
int res = 0;
if (isBST(root, INT_MIN, INT_MAX)) {
return countNodes(root);
}
res = max(res, largestBSTSubtree(root->left));
res = max(res, largestBSTSubtree(root->right));
return res;
}
bool isBST(TreeNode* root, int minVal, int maxVal) {
if (!root) return true;
if (root->val >= maxVal)
return false;
if (root->val <= minVal)
return false;
return isBST(root->left, minVal, root->val) && isBST(root->right, root->val, maxVal);
}
int countNodes(TreeNode* root) {
if (!root) return 0;
return 1 + countNodes(root->left) + countNodes(root->right);
}
private:
int res = 0;
};
日期
2019 年 9 月 20 日 —— 是选择中国互联网式加班?还是外企式养生?
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