Problem-solving ideas:
Verify whether each of the left and right subtree root is a binary tree, while the maximum value is smaller than the left subtree root-> val, the minimum value is greater than the right subtree root-> val.
In the left and right subtrees, has been exploring sub-tree root to the left, you can get its minimum, has the right to explore, you can get the maximum value.
/**
* 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) return true;
return isLeftBST(root->left, root->val) && isRightBST(root->right, root->val);
}
bool isLeftBST(TreeNode* root, int n){
if(isValidBST(root)){
while(root){
if(root->val >= n) return false;
root = root->right;
}
return true;
}
else return false;
}
bool isRightBST(TreeNode* root, int n){
if(isValidBST(root)){
while(root){
if(root->val <= n) return false;
root = root->left;
}
return true;
}
else return false;
}
};