LeetCode || Validate Binary Search Tree

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.

/**
* 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:

//中序遍历的实现
void GetInOrder(TreeNode* root, vector<int>& v)
{
if(root == NULL)
return;

if(root->left!=NULL)
{
GetInOrder(root->left, v);
}
v.push_back(root->val);
if(root->right != NULL)
GetInOrder(root->right, v);
}

bool isValidBST(TreeNode* root) {
//错误的做法：运用递归来实现遍历，避免出现【10， 5， 15， null, null, 6， 20】此类树的结构从而出现错误的判断。
//正确的做法: 利用中序遍历得到的是排好序的结构。

vector<int> v;
GetInOrder(root, v);

for(int i = 1;i<v.size();i++)
{
if(v[i] <= v[i-1])
return false;
}

return true;
}
};