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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Your article gave me a lot of inspiration, I hope you can explain your point of view in more detail, because I have some doubts, thank you.

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