98. Validate Binary Search Tree（中序遍历判断BST）

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.

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分析：

就是判断一个给定的二叉树是否为二叉查找树。

我的思路是：先将该树中序遍历一遍，按中序遍历的顺序保存到一个vector中，然后判断vector中的顺序即可。

ac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:
    vector<int>v;
    bool isValidBST(TreeNode* root) {
        if(root==nullptr)
            return true;
        inorder(root);
        int i,n=v.size();
        for(i=1;i<n;i++)
            if(v[i]<=v[i-1])
            return false;
        return true;
    }
    void inorder(TreeNode* root)
    {
        if(root->left)
            inorder(root->left);
        v.push_back(root->val);
        if(root->right)
            inorder(root->right);
    }
};