[LeetCode] 98. 验证二叉搜索树

题目链接 : https://leetcode-cn.com/problems/validate-binary-search-tree/

题目描述:

给定一个二叉树，判断其是否是一个有效的二叉搜索树。

假设一个二叉搜索树具有如下特征：

• 节点的左子树只包含小于当前节点的数。
• 节点的右子树只包含大于当前节点的数。
• 所有左子树和右子树自身必须也是二叉搜索树。

示例:

示例 1:

输入:
    2
   /   1   3
输出: true
示例 2:

输入:
    5
   /   1   4
     /     3   6
输出: false
解释: 输入为: [5,1,4,null,3,6]。
     根节点的值为 5 ，但是其右子节点值为 4 。

思路:

因为二叉搜索树中序遍历是递增的,所以我们可以中序遍历判断前一数是否小于后一个数.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self,x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isValidBST(self,root: TreeNode) -> bool:
        res = []
        def helper(root):
            if not root:
                return 
            helper(root.left)
            res.append(root.val)
            helper(root.right)
        helper(root)
        return res == sorted(res) and len(set(res)) == len(res)

思路一:迭代

我们可以通过中序遍历迭代方式94. 二叉树的中序遍历来判断.

思路二:递归

1. 中序遍历递归
2. 利用max_val 和min_val

代码:

迭代

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self,root: TreeNode) -> bool:
        stack = []
        p = root
        pre = None
        while p or stack:
            while p:
                stack.append(p)
                p = p.left
            p = stack.pop()
            if pre and p.val <= pre.val:
                return False
            pre = p
            p = p.right
        return True

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        Deque<TreeNode> stack = new LinkedList<>();
        TreeNode p = root;
        TreeNode pre = null;
        while (p != null || !stack.isEmpty()) {
            while (p != null) {
                stack.push(p);
                p = p.left;
            }
            p = stack.pop();
            if (pre != null && pre.val >= p.val) return false;
            pre = p;
            p = p.right;
        }
        return true;
    }
}

思路二

利用递归中序遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self,root: TreeNode) -> bool:
        self.pre = None
        def isBST(root):
            if not root:
                return True
            if not isBST(root.left):
                return False
            if self.pre and self.pre.val >= root.val:
                return False
            self.pre = root
            #print(root.val)
            return  isBST(root.right)
        return isBST(root)

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    TreeNode pre = null;

    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        if (!isValidBST(root.left)) return false;
        if (pre != null && pre.val >= root.val) return false;
        pre = root;
        return isValidBST(root.right);
    }
}

利用最大值最小值

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self,root: TreeNode) -> bool:
        def isBST(root,min_val,max_val):
            if root == None:
                return True
            # print(root.val)
            if root.val >= max_val or root.val <= min_val:
                return False
            return isBST(root.left,root.val) and isBST(root.right,root.val,max_val)
        return isBST(root,float("-inf"),float("inf"))

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        return isBST(root,Long.MAX_VALUE,Long.MIN_VALUE);
    }

    private boolean isBST(TreeNode root,long maxValue,long minValue) {
        if (root == null) return true;
        if (root.val >= maxValue || root.val <= minValue) return false;
        return isBST(root.left,minValue) && isBST(root.right,maxValue,root.val);
    }
}