[LeetCode]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.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:
   1
  / \
 2   3
    /
   4
    \
     5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

思路一：二叉树（二叉搜索树）的中序遍历（左中右）应该是递增的序列。
所以只要写一下中序遍历即可。

/**
 * 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) {
        stack<TreeNode*> sta;
        TreeNode* T=root;
        double left_ele = INT_MIN-1.0;
        while(!sta.empty()||T){
            while(T){
                sta.push(T);
                T=T->left;
            }
            T = sta.top();
            sta.pop();
            if(T->val>left_ele)
                left_ele = T->val;
            else
                return false;
            T =T->right;
        }
        return true;
    }

};

16ms AC

第二种思路就是递归，向左递归时，更新上界；向右递归时，更新下界。

/**
 * 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) {
        double upper = INT_MAX+1.0;
        double lower = INT_MIN-1.0;
        return check(root, upper,lower);

    }
    bool check(TreeNode* T, double upper, double lower){
        if(!T)
            return true;
        else if(T->val>=upper||T->val<=lower)
                return false;
        return check(T->left,T->val,lower)&check(T->right,upper,T->val);

    }


};

16ms AC