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:
2
/ \
1 3
Input: [2,1,3]
Output: true
Example 2:
5
/ \
1 4
/ \
3 6
Input: [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.
My solution
- Do inorder traversal and keep it in stack.
- Iterate through stack to see if any of the value on top is less than equal to second top.
Definition for a binary tree node.
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
class Solution {
public void doInOrderTraversal(TreeNode root, Stack s) {
if(root == null) {
return;
}
doInOrderTraversal(root.left, s);
s.push(root.val);
doInOrderTraversal(root.right, s);
}
public boolean isValidBST(TreeNode root) {
if(root == null) {
return true;
}
Stack<Integer> stack = new Stack<>();
doInOrderTraversal(root, stack);
while(!stack.isEmpty()) {
int top = stack.pop();
if(stack.isEmpty()) {
return true;
}
int secondTop = stack.peek();
if(top <= secondTop) {
return false;
}
}
return true;
}
}
I was thinking to not use stack, but two values current and previous. And keep check if current is less than previous then only break. I am not sure, how to do this. Please suggest.