README.md

98. 验证二叉搜索树

English Version

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 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,null,3,6]。
     根节点的值为 5 ，但是其右子节点值为 4 。

解法

中序遍历，若是一个有效的二叉搜索树，那么遍历到的序列应该是单调递增的。所以只要比较判断遍历到的当前数是否 >= 上一个数即可。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isValidBST(self, root: TreeNode) -> bool:
        def dfs(root):
            nonlocal prev
            if root is None:
                return True
            if not dfs(root.left):
                return False
            if prev >= root.val:
                return False
            prev = root.val
            if not dfs(root.right):
                return False
            return True

        prev = float('-inf')
        return dfs(root)

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private Integer prev;

    public boolean isValidBST(TreeNode root) {
        prev = null;
        return dfs(root);
    }

    private boolean dfs(TreeNode root) {
        if (root == null) {
            return true;
        }
        if (!dfs(root.left)) {
            return false;
        }
        if (prev != null && prev >= root.val) {
            return false;
        }
        prev = root.val;
        if (!dfs(root.right)) {
            return false;
        }
        return true;
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* prev;

    bool isValidBST(TreeNode* root) {
        prev = nullptr;
        return dfs(root);
    }

    bool dfs(TreeNode* root) {
        if (!root) return true;
        if (!dfs(root->left)) return false;
        if (prev && prev->val >= root->val) return false;
        prev = root;
        if (!dfs(root->right)) return false;
        return true;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isValidBST(root *TreeNode) bool {
	var prev *TreeNode

	var dfs func(root *TreeNode) bool
	dfs = func(root *TreeNode) bool {
		if root == nil {
			return true
		}
		if !dfs(root.Left) {
			return false
		}
		if prev != nil && prev.Val >= root.Val {
			return false
		}
		prev = root
		if !dfs(root.Right) {
			return false
		}
		return true
	}

	return dfs(root)
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isValidBST = function (root) {
    let prev = null;

    let dfs = function (root) {
        if (!root) {
            return true;
        }
        if (!dfs(root.left)) {
            return false;
        }
        if (prev && prev.val >= root.val) {
            return false;
        }
        prev = root;
        if (!dfs(root.right)) {
            return false;
        }
        return true;
    };

    return dfs(root);
};

C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    private TreeNode prev;

    public bool IsValidBST(TreeNode root) {
        prev = null;
        return dfs(root);
    }

    private bool dfs(TreeNode root) {
        if (root == null)
        {
            return true;
        }
        if (!dfs(root.left))
        {
            return false;
        }
        if (prev != null && prev.val >= root.val)
        {
            return false;
        }
        prev = root;
        if (!dfs(root.right))
        {
            return false;
        }
        return true;
    }
}

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