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.
Example 1:
2
/
1 3
Binary tree[2,1,3], return true.
Example 2:
1
/ \
2 3
Binary tree[1,2,3], return false.
class Solution {
public:
vector<int> rec;
bool isValidBST(TreeNode* root) {
if(root==NULL)
return true;
if(root->left==NULL&&root->right==NULL)
return true;
inOrder(root);
for(int i=0;i<rec.size()-1;i++)
{
if(rec[i]>=rec[i+1])
return false;
}
return true;
}
void inOrder(TreeNode* root)
{
if(root!=NULL)
{
inOrder(root->left);
rec.push_back(root->val);
inOrder(root->right);
}
}
};
