[LeetCode] 98. 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.

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.

验证二叉搜索树。

题目即是题意。二叉搜索树的特性是对于每个node而言，他的左子树上任意节点都比他自身小，右子树上任意节点都比他自身大。这个题也是有两种做法，迭代和递归。时间空间复杂度都是O(n)。

迭代

迭代的做法会利用到inorder的做法，也是用到一个栈。按照inorder的顺序遍历所有的node，记录一个pre node和一个currrent node。因为是inorder的关系，如果这个BST是有效的，inorder的遍历结果会是有序的。

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isValidBST = function (root) {
    if (root === null) return true;
    let stack = [];
    let cur = root;
    let pre = null;
    while (cur !== null || stack.length) {
        while (cur !== null) {
            stack.push(cur);
            cur = cur.left;
        }
        cur = stack.pop();
        if (pre !== null && cur.val <= pre.val) {
            return false;
        }
        pre = cur;
        cur = cur.right;
    }
    return true;
};

Java实现

class Solution {
    public boolean isValidBST(TreeNode root) {
        // corner case
        if (root == null) {
            return true;
        }

        // normal case
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        TreeNode pre = null;
        while (cur != null || !stack.isEmpty()) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            if (pre != null && cur.val <= pre.val) {
                return false;
            }
            pre = cur;
            cur = cur.right;
        }
        return true;
    }
}

递归

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isValidBST = function (root) {
    if (root === null) return true;
    return helper(root, null, null);
};

var helper = function (root, min, max) {
    if (root === null) return true;
    if (min !== null && root.val <= min) return false;
    if (max !== null && root.val >= max) return false;
    return helper(root.left, min, root.val) && helper(root.right, root.val, max);
}

[更新] 另一种递归

利用inorder递归，再判断数组是否有序。

时间O(n)

空间O(n) - 递归的栈空间O(n) + 存放节点的数组O(n)

/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isValidBST = function (root) {
    let res = [];
    if (root === null) return true;
    inorder(res, root);
    for (let i = 1; i < res.length; i++) {
        if (res[i] <= res[i - 1]) return false;
    }
    return true;
};

var inorder = function (res, root) {
    if (root === null) return true;
    inorder(res, root.left);
    res.push(root.val);
    inorder(res, root.right);
}

Java实现

class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        return helper(root, null, null);
    }

    private boolean helper(TreeNode root, Integer min, Integer max) {
        if (root == null) return true;
        if (min != null && root.val <= min) return false;
        if (max != null && root.val >= max) return false;
        return helper(root.left, min, root.val) && helper(root.right, root.val, max);
    }
}

LeetCode 题目总结