leetcode 98. 验证二叉搜索树 Validate Binary Search Tree （使用c++/java/python）

https://leetcode-cn.com/problems/validate-binary-search-tree/

https://leetcode.com/problems/validate-binary-search-tree/

利用一个函数判断子树是否在上下限之间，当上下限为null时表示无限大。

每递归一次，父节点就变成左子树的上限、右子树的下限。

执行用时： c++ 8ms； java 1ms； python 84ms

c++

/**
 * 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 isValidSubTree(root,NULL,NULL);
    }
    bool isValidSubTree(TreeNode* root, TreeNode* lowerBound, TreeNode* upperBound)
    {
        // 判断子树是否在上界和下界之间
        if(root==NULL) return true;
        if( (lowerBound && lowerBound->val >= root->val) || (upperBound && upperBound->val <= root->val) )
            // 下界存在时做判断、上界存在时做判断
            return false;
        return isValidSubTree(root->left,lowerBound,root)&&isValidSubTree(root->right,root,upperBound);
        //父节点作为左子树的上界、右子树的下界
    }
};

 java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        return isValidSubBST(root,null,null);
    }
    public boolean isValidSubBST(TreeNode root, TreeNode lowerB, TreeNode upperB)
    {
        if(root==null)
            return true;
        if((lowerB!=null && root.val <= lowerB.val)||(upperB!=null && root.val >= upperB.val))
            return false;
        return isValidSubBST(root.left,lowerB,root)&&isValidSubBST(root.right,root,upperB);
    }
}

python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        return self.isValidSubTree(root,None,None)
    
    def isValidSubTree(self, root, lowerB, upperB):
        if root is None:
            return True
        if (lowerB and lowerB.val>=root.val) or (upperB and upperB.val<=root.val):
            return False
        return self.isValidSubTree(root.left,lowerB,root) and self.isValidSubTree(root.right,root,upperB)