Problem
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.
Example1
Example2
Solution
利用BST中序遍是递增序列的特性。
非递归,中序遍历BST。
/**
* 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:
bool isValidBST(TreeNode* root) {
if(!root)
return true;
stack<TreeNode*> sk;
TreeNode *cur = root;
TreeNode *prev = NULL;
while(cur || !sk.empty())
{
while(cur)
{
sk.push(cur);
cur = cur->left;
}
cur = sk.top();
sk.pop();
if(prev)
{
if(prev->val >= cur->val)
{
return false;
}
else
{
prev = cur;
}
}
else
{
prev = cur;
}
cur = cur->right;
}
return true;
}
};