LeetCode 98. Validate Binary Search Tree

原题链接在这里：https://leetcode.com/problems/validate-binary-search-tree/

题目： 

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. 

题解：

根据BST特性递归调用原函数，如果出现root.val >= max 或者root.val <= min的情况return false. 

Note: 初始的max 和 min 要设成long 型的最大最小值，因为边界的corner case. 若是用Integer.MAX_VALUE. 并且数只有一个节点，刚开始若root.val = Integer.MAX_VALUE, 会返回false. 其实应该返回true.

Time Complexity: O(n). Space: O(logn).

AC Java:

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

AC C++:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        return isValidDfs(root, LONG_MIN, LONG_MAX);
    }

private:
    bool isValidDfs(TreeNode* root, long min, long max){
        if(!root){
            return true;
        }

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

        return isValidDfs(root->left, min, root->val) && isValidDfs(root->right, root->val, max);
    }
};

另外一种方法是做inorder traverse, 若是出现逆序，就返回false.

Time Complexity: O(n). Space: O(logn).

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

类似Find Mode in Binary Search Tree, Minimum Absolute Difference in BST.