98 Validate Binary Search Tree

98. Validate Binary Search Tree

1. Question

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

Binary tree[2,1,3], return true.

Example 2:

    1
   / \
  2   3

Binary tree[1,2,3], return false.

2. Implementation

(1) Inorder Traversal

class Solution {
    public boolean isValidBST(TreeNode root) {
        TreeNode curNode = root, preNode = null;
        Stack<TreeNode> stack = new Stack<>();

        while (curNode != null || !stack.isEmpty()) {
            if (curNode != null) {
                stack.push(curNode);
                curNode = curNode.left;
            }
            else {
                curNode = stack.pop();
                if (preNode != null && preNode.val >= curNode.val) {
                    return false;
                }
                preNode = curNode;
                curNode = curNode.right;
            }
        }
        return true;
    }
}

(2) Binary Search Tree Property

class Solution {
    public boolean isValidBST(TreeNode root) {
        return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }

    public boolean isValidBST(TreeNode node, long min, long max) {
        if (node == null) {
            return true;
        }

        if (node.val <= min || node.val >= max) {
            return false;
        }
        return isValidBST(node.left, min, node.val) && isValidBST(node.right, node.val, max);
    }
}

3. Time & Space Complexity

Inorder Traversal: 时间复杂度:O(n), 空间复杂度: O(h)

Binary Search Tree Property: 时间复杂度: O(n), 空间复杂度: O(h)