Given the root
of a binary tree, determine if it is a valid binary search tree (BST).
A valid 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:

Input: root = [2,1,3] Output: true
Solution
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> ans;
queue<TreeNode*> q;
if(root)
q.push(root);
int i = 0;
while(!q.empty()){
vector<int> temp;
int n = q.size();
for(int i=0;i<n;i++){
TreeNode* _node = q.front();
q.pop();
temp.push_back(_node->val);
if(_node->left)
q.push(_node->left);
if(_node->right)
q.push(_node->right);
}
ans.push_back(temp);
}
return ans;
}
};
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