98. Validate Binary Search Tree
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
- The left subtree of a node contains only nodes with keys less than the node's key.
- The right subtree of a node contains only nodes with keys greater than the node's key.
- Both the left and right subtrees must also be binary search trees.
Example 1:
2
/ \
1 3
Input: [2,1,3]
Output: true
Example 2:
5
/ \
1 4
/ \
3 6
Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.
题目链接:https://leetcode-cn.com/problems/validate-binary-search-tree/
法一:递归
递归的写法也有多种,除了这里写到的可以继续思考有没有更优美的写法。
坑:
1)上下边界都要记录,因为还在节点有左有右,判定边界有区别。
2)如果把上下边界初始化为INT_MAX和INT_MIN,要考虑元素值等于边界的情况,以及更新后的元素值又等于INT_MAX和INT_MIN的情况。
/**
* 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:
bool isValidBST(TreeNode* root) {
return validate(root, LONG_MAX, LONG_MIN);
}
bool validate(TreeNode* root, long high, long low){
if(!root || (!root->left && !root->right)) return true;
auto l = root->left, r = root->right;
int mid = root->val;
if(l && (l->val>=mid || l->val<=low)) return false;
if(r && (r->val<=mid || r->val>=high)) return false;
return validate(l, mid, low) && validate(r, high, mid);
}
};
法二:中序遍历
将树中序遍历输出,如果输出的不是严格递增数列,则false。