Leetcode[98]-Validate Binary Search Tree

Link: https://leetcode.com/problems/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.
confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.

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思路：递归方法，对于根结点

• 如果有左子树，比较根结点与左子树的最大值，如果小于等于则返回false；
• 如果有右子树，比较根结点与右子树的最小值 ，如果大于等于则返回false；
• 接着判断左子树和右子树是否也是合法的二叉搜索树；

Code(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) {        if(root == NULL) return true;        if(root->left && root->val <= leftMax(root->left)) return false;        if(root->right && root->val >= rightMin(root->right)) return false;        return isValidBST(root->left) && isValidBST(root->right);    }    //get the left tree max value    int leftMax(TreeNode *root){        if(root == NULL) return INT_MIN;        int max = root->val;        int lmax = leftMax(root->left);        int rmax = leftMax(root->right);        int max1 = lmax>rmax?lmax:rmax;        return max>max1?max:max1;    }    //get the right tree minimum value    int rightMin(TreeNode *root){        if(root == NULL) return INT_MAX;        int max = root->val;        int lmin = rightMin(root->left);        int rmin = rightMin(root->right);        int max1 = lmin>rmin? rmin:lmin;        return max>max1?max1:max;    }};