leetcode Validate Binary Search Tree

Validate Binary Search Tree

Total Accepted: 3218 Total
Submissions: 12801My Submissions

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.

confused what

"{1,#,2,3}"

means? >
read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as

"{1,2,3,#,#,4,#,#,5}"

.

Take notice that

The left subtree of a node contains only nodes with keys less
than the node's key

I.e., all nodes are less than the root.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
 public:
  bool check(TreeNode *root, int min, int max) {
    if (!root)
      return true;
    if (root->val <= min || root->val >= max)
      return false;
    if (root->left && !check(root->left, min, root->val))
      return false;
    if (root->right && !check(root->right, root->val, max))
      return false;
    return true;
  }
  bool isValidBST(TreeNode *root) {
    if (!root)
      return true;
    return check(root, INT_MIN, INT_MAX);
  }
};

Another method is to give an inorder traversal of Binary Tree. For example,

/**
 * Definition for binary tree
 * struct TreeNode {
 * int val;
 * TreeNode *left;
 * TreeNode *right;
 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        //注意题目要求是 less than和greater than;
        stack<TreeNode*> S;
        TreeNode *pre = NULL, *p = root;
        while(p || S.empty() == false)
        {
            while(p)
            {
                S.push(p);
                p = p->left;
            }
            if(S.empty() == false)
            {
                p = S.top();
                S.pop();
                if(pre && p->val <= pre->val)return false;
                pre = p;
                p = p->right;
            }
        }
        return true;
    }
};