Day 45, #98 Validate Binary Search Tree

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.

---------------------------------------------------
Perform an in-order tree traversal, save all values in an array, and later check if it is sorted
There is room for space optimization in solution below

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void rec (TreeNode *root, vector<int> &v) {
        if (root->left != NULL) {
            rec(root->left,v);
        }
        v.push_back(root->val);
        if (root->right != NULL) {
            rec(root->right,v);
        }
    }

    bool isValidBST(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root == NULL) return true; 
        vector<int> v;
        rec(root,v);
        if (v.size() < 2) {
            return true;
        }
        // check if it's already sorted
        for (int i = 1; i < v.size(); i++) {
            if (v[i-1] >= v[i]) {
                return false;
            }
        }
        return true;
    }
};

Update: Jan-16-2014
http://leetcode.com/2010/09/determine-if-binary-tree-is-binary.html
Solution#2, in space.
Note int &pre in arguments

/**
 * 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 inorder (TreeNode *root, int &pre) {
       if (root == NULL) {
           return true;
       }
       if (inorder(root->left,pre)) {
           if (pre >= root->val) {
               return false;
           }
           pre = root->val;
           return inorder(root->right,pre);
       }
       return false;
       
    }

    bool isValidBST(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root == NULL) return true; 
        int pre = INT_MIN;
        return inorder(root,pre);
    }
};

Update: Jan-15-2015
to handle INT_MIN case

/**
 * 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 inorder(TreeNode *root, TreeNode *&pre) {
        if (root == NULL) {
            return true;
        }
        
        if (inorder(root->left,pre)) {
            if (pre != NULL && pre->val >= root->val) {
                return false;
            }
            pre = root;
            return inorder(root->right,pre);
            
        }
        
        return false;
    }

    bool isValidBST(TreeNode *root) {
        TreeNode *pre = NULL;
        return inorder(root,pre);
    }
};

Java, updated Jun-24th-2018

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    private TreeNode pre = null;
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        if (!isValidBST(root.left)) {
            return false;
        }
        
        if (pre == null) pre = root;
        else if (pre.val >= root.val) {
            return false;
        }
        pre = root;
        return isValidBST(root.right);
    }
}

当有node的值等于MIN_VALUE或MAX_VALUE, 这算法就搞不定了。所有还是别用了

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        return helper(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
    }
    
    private boolean helper(TreeNode root, int min, int max) {
        if (root == null) return true;
        if (root.val < max && root.val > min) {
            return helper(root.left, min, root.val) 
                && helper(root.right, root.val, max);
        }
        
        return false;
    }
}