Leetcode | 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.

Method I

这道题应该在Recover Binary Search Tree之前做。

中序遍历，记得保存前一个节点。只要pre->val>=root->val就错了。

 1 class Solution {

 2 public:

 3     bool isValidBST(TreeNode *root) {

 4         pre = NULL;

 5         return inorder(root);

 6     }

 7     

 8     bool inorder(TreeNode *root) {

 9         if (!root) return true;

10         if (!inorder(root->left)) return false;

11         if (pre && pre->val >= root->val) return false;

12         pre = root;

13         return inorder(root->right);

14     }

15 private:

16     TreeNode* pre;

17 };

用Morris Traversal的方式竟然报runtime error。同样的代码可以跑在本机上，修改一下也可以通过Recover Binary Search Tree。真是奇怪。难道是leetccode上这道题不允许修改树本身。

 1 class Solution {

 2 public:

 3     bool isValidBST(TreeNode *root) {

 4         TreeNode* pre = NULL;

 5         TreeNode* current = root;

 6         

 7         while (current != NULL) {

 8             if (current->left == NULL) {

 9                 if (pre != NULL && pre->val >= current->val) return false;

10                 pre = current;

11                 current = current->right;

12             } else {

13                 TreeNode* rightmost = current->left;

14                 while (rightmost->right != NULL && rightmost->right != current) rightmost = rightmost->right;

15                 if (rightmost->right == NULL) {

16                     rightmost->right = current;

17                     current = current->left;

18                 } else {

19                     rightmost->right = NULL;

20                     if (pre != NULL && pre->val >= current->val) return false;

21                     pre = current;

22                     current = current->right;

23                 }

24             }

25         }

26         return true;

27     }

28 };

 Method II

参考自career cup chapter 4.

维护一个区间(min, max)，只要左子树、右子树都在一个正确的区间里就行了。

在检查左子树时，区间为(min, root->val)；

检查右子树时，区间为(root->val, max)；

注意是开区间。

 1 /**

 2  * Definition for binary tree

 3  * struct TreeNode {

 4  *     int val;

 5  *     TreeNode *left;

 6  *     TreeNode *right;

 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 8  * };

 9  */

10 class Solution {

11 public:

12     bool isValidBST(TreeNode *root) {

13         return recursive(root, INT_MIN, INT_MAX);

14     }

15     

16     bool recursive(TreeNode *root, int min, int max) {

17         if (root == NULL) return true;

18         if (root->val <= min || root->val >= max) return false;

19         return recursive(root->left, min, root->val) && recursive(root->right, root->val, max);

20     }

21 };