【LeetCode】98. Validate Binary Search Tree (2 solutions)

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.

 

解法一： 

利用二叉搜索树的特点，中序遍历然后判断是否严格升序。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode *root) {
        vector<int> ret;
        inOrder(root, ret);
        if(ret.size() <= 1)
            return true;
        for (int i = 1; i < ret.size(); i ++)
        {
            if(ret[i] <= ret[i-1])
                return false;
        }
        return true;
    }
    void inOrder(TreeNode *root, vector<int>& ret)
    {
        if(!root)
            return;
        if(root->left)
            inOrder(root->left, ret);
        ret.push_back(root->val);
        if(root->right)
            inOrder(root->right, ret);
    }
};

 

解法二：

由于只需要关心前一个节点与当前的大小关系，因此不用保存所有值。

只需要使用一个全局变量保存前一个节点值即可。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution 
{
public:
    TreeNode* last = NULL;
    bool isValidBST(TreeNode *root) 
    {
        if(root == NULL)
            return true;
        else
        {
            bool leftRes = isValidBST(root->left);
            //short cut
            if(leftRes == false)
                return false;
            if(last && last->val >= root->val)
                return false;
            last = root;
            return isValidBST(root->right);
        }
    }
};