# 98. Validate Binary Search Tree (Medium)

https://leetcode.com/problems/validate-binary-search-tree/

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:

```Input:
2
/ \
1   3
Output: true
```

Example 2:

```    5
/ \
1   4
/ \
3   6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
is 5 but its right child's value is 4.
```

## Solutions

``````class Solution {

// Actually, the best solution is to traverse the tree and return the visiting sequence.
// Then sort out if the outcome is sorted by ascending order or not. If not well ordered,
// this tree must not be the valid Binary Sort Tree.

public boolean isValidBST(TreeNode root) {
List<Integer> track = new ArrayList<>();

traverse(root, track);

for (int i = 1; i < track.size(); i++) {
if (track.get(i - 1) >= track.get(i)) {
return false;
}
}

return true;
}

private void traverse(TreeNode root, List<Integer> track) {
if (root == null) {
return;
}

traverse(root.left, track);

traverse(root.right, track);
}
}
``````

## Incorrect Solutions

``````class Solution2 {

// This solution is trivial, nasty and incorrect.

// You can look into this test case: [3,1,5,0,2,4,6,null,null,null,3] and will
// find out what's going wrong here.

public boolean isValidBST(TreeNode root) {
if (root == null) {
return true;
}

return check(root, "", root.val);
}

private boolean check(TreeNode root, String from, int limit) {
if (root == null) {
return true;
}

// 1 Make sure the value order is left < root < right = is also not permitted
if (root.left != null && root.left.val >= root.val) {
return false;
}

if (root.right != null && root.right.val <= root.val) {
return false;
}

// 2.1 If this branch is the left one of parent, this.right.val < parent.val;
if (from.equals("left") && root.right != null && root.right.val >= limit) {
return false;
}

// 2.2 If this branch is the right one of parent, this.left.val > parent.val;
if (from.equals("right") && root.left != null && root.left.val <= limit) {
return false;
}

// If left branch is invalid, not need to validate right branch, return immediately.
if (!check(root.left, "left", root.val)) {
return false;
}

if (!check(root.right, "right", root.val)) {
return false;
}

return true;
}
}
``````