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.
合法查找二叉树的检验。
要求所有左节点值比自身小，所有右节点值比自身大。
因此在每次返回的函数中，节点N的左节点应大于N应该大于的值&&小于N，右节点应大于N&&小于N应该小于的值。
/**
* 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 isValidBST(root, NULL, NULL);
}
bool isValidBST(TreeNode* root, TreeNode* minNode, TreeNode* maxNode) {
if(!root) return true;
if(minNode && root->val <= minNode->val || maxNode && root->val >= maxNode->val)
return false;
return isValidBST(root->left, minNode, root) && isValidBST(root->right, root, maxNode);
}
};
