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 };