# Leetcode Validate-Binary-Search-Tree

#### 2022-07-08 10:36

> ##### Algorithms:
>
> #algorithm #DFS #DFS_inorder
>
> ##### Data structures:
>
> #DS #binary_tree #binary_search_tree
>
> ##### Difficulty:
>
> #coding_problem #difficulty-medium
>
> ##### Additional tags:
>
> #leetcode #CS_list_need_practicing
>
> ##### Revisions:
>
> N/A

##### Related topics:

##### Links:

- [Link to problem](https://leetcode.com/problems/validate-binary-search-tree/)

---

### Problem

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.

#### Examples

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/12/01/tree1.jpg)

**Input:** root = [2,1,3]
**Output:** true

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/12/01/tree2.jpg)

**Input:** root = [5,1,4,null,null,3,6]
**Output:** false
**Explanation:** The root node's value is 5 but its right child's value is 4.

#### Constraints

- The number of nodes in the tree is in the range `[1, 104]`.
- `-231 <= Node.val <= 231 - 1`

### Thoughts

> [!summary]
> This is a #DFS #DFS_inorder problem.

I have thought a lot of recursion methods, but at last I realized:

> [!tip] The feature of BST
> For a BST, DFS inorder search returns an array of ascending numbers.

So, I use a DFS inorder, along with a prev reference pointer to keep track of previous value. to validate, the value of node should always be bigger than prev.

See comment for why I use a **reference to pointer.**

### Solution

```cpp
/**
 * 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 {
  bool checker(TreeNode *root, int *&prev) {
    // Use reference to pointer here, because the pointer address will change
    // when assigning memory. Also can use pointer to pointer, or initialize the
    // pointer and never change the address ever.
    if (!root) {
      return true;
    }
    if (!checker(root->left, prev)) {
      return false;
    }
    if (prev && root->val <= *prev) {
      return false;
    }
    if (!prev) {
      // prev's address got changed
      prev = new int;
    }
    *prev = root->val;
    return checker(root->right, prev);
  }

public:
  bool isValidBST(TreeNode *root) {
    // DFS inorder traversal: valid BST returns a ascending array
    int *prev = nullptr;
    return checker(root, prev);
  }
};
```
