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() <= )
return true;
for (int i = ; i < ret.size(); i ++)
{
if(ret[i] <= ret[i-])
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);
}
}
};