LeetCode [98] Validate Binary Search Tree

98. Validate Binary Search Tree

Medium

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.

 

Example 1:

    2
   / \
  1   3

Input: [2,1,3]
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6

Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

//C++: method2 22ms
typedef long long ll;
class Solution {
public:
    bool inorder(TreeNode *root, ll &pre){
        if(!root) return true;
        if(!inorder(root->left, pre)) return false;
        if(pre>=root->val){ 
            return false;
        }
        pre = root->val;
        return inorder(root->right, pre);
    }
    
    bool isValidBST(TreeNode *root) {
        ll pre = -2147483649;
        return inorder(root, pre);
    }
};
]]></script>
<script class="brush: js" type="syntaxhighlighter"><![CDATA[
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        stack<TreeNode *> stk;
        long long lb = (long long)(INT_MIN)-1;
        TreeNode* p = root;
        while(p){
            stk.push(p);
            p = p->left;
        }
        while(stk.size()){
            p = stk.top();
            stk.pop();
            if(lb>=p->val) return false;
            lb = p->val;
            if(p->right){
                p = p->right;
                while(p){
                    stk.push(p);
                    p = p->left;                    
                }
            }
        }
        return true;  
    }
};

//Java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        long[] ret = valid(root);
        return (int)ret[0]==1;
    }
    
    long[] valid(TreeNode node){
        if(node == null) return new long[]{1, Long.MAX_VALUE, Long.MIN_VALUE};
        long[] l = valid(node.left);
        long[] r = valid(node.right);
        boolean isBST = (int)l[0]==1 && (int)r[0]==1 && l[2]<(long)node.val && node.val<(long)r[1];
        return new long[]{isBST?1:0, Long.min(l[1], (long)node.val), Long.max((long)node.val, r[2])};
    }
}