98. Validate Binary Search Tree
problem description
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:
2
/ \
1 3
Input: [2,1,3]
Output: true
Example 2:
5
/ \
1 4
/ \
3 6
Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.
algorithm thought
得到合法的BST,BST有个性质就是中序遍历结果是一个有序数组。所以我们直接中序遍历,每次保存前一个值,如果前一个值大于等于当前遍历的值,那么就不是一个合法的BST,否则就是合法的BST
code
/**
* 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) {
stack<TreeNode*> todo;
long pre=-2147483649;
while(root||!todo.empty()){
while(root){
todo.push(root);
root=root->left;
}
root=todo.top();todo.pop();
if(root->val<=pre){
return false;
}
pre=root->val;
root=root->right;
}
return true;
}
};
algorithm analysis
时间复杂度O(n),遍历的时间,空间复杂度O(n),使用一个栈保存数据。其实可以用之前的Morris tarversal但是LeetCode可能是有bug,代码是正确的但是最后得不到合理的结果。
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