Validate Binary Search Tree

问题描述

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.

 

解决思路

1. 递归：当前节点的值 与 左节点的最右子节点（左子树最大的节点）、右节点的最左子节点（右子树最小的节点） 比较；

2. 非递归中序遍历（前后指针）；

方法1较方法2有较多的重复计算，方法2的时间复杂度为O(n).

 

程序

1. 递归

public class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }    
        
        long leftMax = getLeftMax(root.left);
        long rightMin = getRightMin(root.right);
        if (leftMax >= root.val || rightMin <= root.val) {
            return false;
        }
        
        return isValidBST(root.left) && isValidBST(root.right);
    }
    
    private long getLeftMax(TreeNode root) {
        if (root == null) {
            return Long.MIN_VALUE;
        }
        TreeNode node = root;
        while (node.right != null) {
            node = node.right;
        }
        return (long)node.val;
    }
    
    private long getRightMin(TreeNode root) {
        if (root == null) {
            return Long.MAX_VALUE;
        }
        TreeNode node = root;
        while (node.left != null) {
            node = node.left;
        }
        return (long)node.val;
    }
}

2. 非递归中序遍历

public class Solution {
    // inorder traversal, two pointers
    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }    
        Stack<TreeNode> s = new Stack<>();
        TreeNode node = root;
        TreeNode pre = null;
        
        while (node != null || !s.isEmpty()) {
            while (node != null) {
                s.push(node);
                node = node.left;
            }
            if (!s.isEmpty()) {
                TreeNode cur = s.pop();
                if (pre != null && pre.val >= cur.val) {
                    return false;
                }
                pre = cur;
                node = cur.right;
            }
        }
        
        return true;
    }
}