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struct TreeNode {	int val;	TreeNode *left;	TreeNode *right;	TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {//hint: when involved with BST, always remember that special effect of in-order traversal//1. pass down the low and high limits from node to node//2. we can always do an in-order traversal of BST, then the value we visited should be in increasing order,//so we can check the previous is smaller than current valuepublic:	bool isBSTHelper(TreeNode* root, int low, int high)	{		if(!root)			return true;		if (low < root->val && high > root->val)			return isBSTHelper(root->left, low, root->val) 			       && isBSTHelper(root->right, root->val, high);		else return false;	}	bool isBSTHelper(TreeNode* root, int& prev)	{		if(!root)			return true;		return isBSTHelper(root->left, prev) && (root->val > prev)				&& ((prev = root->val)||true) && isBSTHelper(root->right, prev);			}	bool isValidBST(TreeNode *root) {		// Start typing your C/C++ solution below		// DO NOT write int main() function		//first solution		//return isBSTHelper(root, INT_MIN, INT_MAX);		//second solution		int prev = INT_MIN;		return isBSTHelper(root, prev);	}};

second time

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isValidBSTUtil(TreeNode* root, int mmin, int mmax)    {        if(root == NULL) return true;        if(mmin < root->val && root->val < mmax)        {            return isValidBSTUtil(root->left, mmin, min(mmax, root->val))                    && isValidBSTUtil(root->right, max(mmin, root->val), mmax);        }        else return false;    }    bool isValidBST(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int mmin = INT_MIN;        int mmax = INT_MAX;        return isValidBSTUtil(root, mmin, mmax);    }};

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