[Leetcode] 98. Validate Binary Search Tree 解题报告

来源：互联网发布：go桌面软件下载编辑：程序博客网时间：2024/05/16 15:15

题目：

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.

思路：

树的问题首先想到递归，一个巧妙的方法就是给当前根节点设定一个合法的范围[min_value, max_value]。那么判断一个树是否为BST的充要条件就是：1）根节点的范围处于[min_value, max_value]之内；2）所有左子树上的节点的值必须处在[min_value, root_value - 1]之内；3）所有右子树上的节点的值必须处在[root_value + 1, max_value]之内。后两个条件可以通过递归实现。

本题还需要注意的实现细节是overflow，为此我们可以将min_value和max_value的类型设为long long，这样可以有效避免int类型的加减运算的溢出问题。

代码：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isValidBST(TreeNode* root) {        return isValidBST(root, INT_MIN, INT_MAX);    }private:    bool isValidBST(TreeNode *root, long long min_value, long long max_value) {        if (root == NULL) {                                     // empty tree is always valid            return true;        }        if (min_value > max_value) {                            // empty range for the root's value            return false;        }        if (root->val < min_value || root->val > max_value) {   // out of the valid range            return false;        }        long long value = static_cast<long long>(root->val);    // in order to avoid overflow        if (!isValidBST(root->left, min_value, value - 1)) {    // check left subtree            return false;        }        if (!isValidBST(root->right, value + 1, max_value)) {   // check right subtree            return false;        }        return true;    }};

0 0