【leetcode】【medium】98. Validate Binary Search Tree

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。