CODE 35: 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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as  "{1,2,3,#,#,4,#,#,5}".

	public boolean isValidBST(TreeNode root) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if (null == root) {
			return true;
		}
		Stack stack = new Stack();
		TreeNode pre = new TreeNode(Integer.MIN_VALUE);
		TreeNode now = root;
		while (!stack.isEmpty() || null != now) {
			if (null == now) {
				now = stack.pop();
				if (pre.val >= now.val) {
					return false;
				}
				pre = now;
				now = now.right;
			} else {
				stack.push(now);
				now = now.left;
			}
		}
		return true;
	}