Validate Binary Search Tree

Submission Detail

82 / 82 test cases passed.
 Status:

Accepted
Runtime: 9 ms
Memory Usage: 44.5 MB
Submitted: 0 minutes ago
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Runtime Error Message:
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Input:
Output:
Expected:

Accepted Solutions Runtime Distribution

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Your runtime beats 5.75 % of java submissions.
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Accepted Solutions Memory Distribution

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Submitted Code: 0 minutes ago

Language: java

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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
boolean BST = true;
return checkTree(root,BST);
}
public boolean checkTree(TreeNode r, boolean BST) {
if (r == null || BST == false)
return BST;
if (r.left != null)
if (r.left.val >= r.val)
BST = false;
if (r.right != null)
if (r.right.val <= r.val)
BST = false;
ArrayList<Integer> inorder = new ArrayList<>();
inorder = inOrderTraversal(r,inorder);
int j = 0;
while (inorder.get(j) != r.val) {
if (inorder.get(j) >= r.val)
return false;
j++;
}
j++;
while (j < inorder.size()) {
if (inorder.get(j) <= r.val)
return false;
j++;
}
BST = checkTree(r.left, BST);
BST = checkTree(r.right, BST);
return BST;
}
public ArrayList<Integer> inOrderTraversal(TreeNode root, ArrayList<Integer> iot) {
if (root == null)
return iot;
iot = inOrderTraversal(root.left,iot);
iot.add(root.val);
iot = inOrderTraversal(root.right,iot);
return iot;
}
}
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