LeetCode 98 Validate Binary Search Tree判断是否为合法二叉树

/**
* 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 Left(TreeNode* left, int root){
while(left->right != NULL){
left = left->right;
}
return left->val < root;
}
bool Right(TreeNode* right,int root){
while(right->left != NULL)
right = right->left;
return right->val > root;
}

bool isValidBST(TreeNode* root) {
if(root == NULL) return true;
int left = 0, right = 0;
bool lres = true,rres = true;
if(root->left){//不为空的情况下
left = 1;
if(root->left->val >= root->val) return false;
lres = isValidBST(root->left);
if(lres) lres = Left(root->left,root->val);//在左儿子是合法二叉树的情况下，检验左儿子的最大值（即一直取左儿子的右儿子）是否小于root值
else return false;
}
if(root->right){
right = 1;
if(root->right->val <= root->val) return false;
rres = isValidBST(root->right);
if(rres) rres = Right(root->right,root->val);
else return false;
}
return lres && rres;
}

};