[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.

分析：二叉搜索树的中序遍历是从小到大排列的。

/**
* 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
{
private:
int pre;
vector<int> vec;
public:
Solution():pre(LONG_MIN) {}

bool isValidBST(TreeNode* root)
{
vec.clear();
if(root == NULL) return true;
if(root->left==NULL && root->right==NULL) return true;
isValid(root);

for(int i=1; i<vec.size(); i++)
if(vec[i-1] >= vec[i])
return false;

return true;
}

void isValid(TreeNode* root)
{
if(root)
{
isValid(root->left);
vec.push_back(root->val);
isValid(root->right);
}
}
};

方法二：设置上下界

class Solution
{
public:
bool isValidBST(TreeNode* root)
{
if(root == NULL) return true;
if(root->left==NULL && root->right==NULL) return true;

return isValid(root, LONG_MIN, LONG_MAX);
}

bool isValid(TreeNode* root, int min, int max)
{
if(root==NULL) return true;
if(root->val <= min || root->val >= max) return false;

return isValid(root->left, min, root->val) && isValid(root->right, root->val, max);
}
};