98. 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.

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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【解法一】我的解法，引入最大最小值

class Solution {public:    bool isValidBST(TreeNode* root) {        return isvalid(root, LONG_MIN, LONG_MAX);    }    bool isvalid(TreeNode *root, long MIN, long MAX) {        if (!root) return true;        if (root->val <= MIN || root->val >= MAX) return false;        return isvalid(root->left, MIN, root->val) && isvalid(root->right, root->val, MAX);    }};

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【解法二】网上的其他解法,中序遍历存入数组中，判断a[i]>a[i+1]

class Solution {public:    vector<int> temp;    bool isValidBST(TreeNode* root) {        inOrderTraversal(root);        if(temp.size()<=1) return true;        for(int i=1;i<temp.size();i++){            if(temp[i]<=temp[i-1]) return false;        }        return true;    }    void inOrderTraversal(TreeNode * root){        if(!root) return;        inOrderTraversal(root->left);        temp.push_back(root->val);        inOrderTraversal(root->right);    }};