[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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

经典binary tree的题目。用一个区间，初始化为(min, max)。依次用它检查树中的每一点。如果在区间中，就是合理的。否则，返回false。对区间的更新遵守这样的原则：如果是左子树的递归，就更新max为root.val；如果是右子树递归，就更新min为root.val。当然，也可以对树进行中序遍历，然后检查得到的遍历结果是不是有序的。但是这样还需要额外的空间，肯定没有用区间的方法好，就不再赘述了。

具体代码如下：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBSTHelper(TreeNode root, int lowerBound, int upperBound) {
        if(root==null) return true;
        if(root.val>lowerBound && root.val<upperBound) {
            return isValidBSTHelper(root.left, lowerBound, root.val) && isValidBSTHelper(root.right, root.val, upperBound);
        }
        else return false;
    }
    
    public boolean isValidBST(TreeNode root) {
        return isValidBSTHelper(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
    }
}

Version 2: Construct an array through in-order traversal. Determine whether this array is sorted or not. 

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void isValidBST(TreeNode root, ArrayList<Integer> inorder) {
        if(root==null) return;
        isValidBST(root.left, inorder);
        inorder.add(root.val);
        isValidBST(root.right, inorder);
        return;
    }
    
    public boolean isValidBST(TreeNode root) {
        ArrayList<Integer> inorder = new ArrayList<Integer>();
        isValidBST(root, inorder);
        for(int i=0; i<inorder.size()-1; i++) {
            if(inorder.get(i)>=inorder.get(i+1)) return false;
        }
        return true;
    }
}