【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
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.
Example 1:
Input:
2
/
1 3
Output: true
Example 2:
5
/
1 4
/
3 6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
is 5 but its right child's value is 4.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
//用中序遍历迭代法做 （递归法没看懂）
//二叉查找树性质：中序遍历后，二叉查找树为升序排列
class Solution
{
public:
bool isValidBST(TreeNode* root)
{
if(root == NULL) return true;
stack<TreeNode*> s;
TreeNode* p = root;
TreeNode* pre = NULL;
while(!s.empty() || p)
{
if(p)
{
s.push(p);
p = p->left;
}
else
{
p = s.top();
if(pre != NULL && p->val <= pre->val) return false; //看是否为升序
pre = p; //保存已经访问的结点
s.pop();
p = p->right;
}
}
return true;
}
};
//参考：
while(p != nullptr || !s.empty())
{
if(p != nullptr) //当左结点不为空时
{
s.push(p); //入栈
p = p->left; //指向下一个左结点
}
else //当左结点为空时
{
p = s.top();
path.push_back(p->val); //访问栈顶元素（父结点）
s.pop(); //出栈
p = p->right; //指向右结点
}
}