Leetcode - Validate Binary Search Tree-优快云博客

本文介绍了一种通过递归检查二叉树是否为有效的二叉搜索树的方法，并详细解释了如何利用节点值域来判断每个节点及其子树是否符合二叉搜索树的定义。

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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.

[分析] 这题较一年之前的版本多了些边界值的test case，看出leetcode在不断优化完善，赞!
对我来说，对边界条件的考虑是bug free的关键。基本思路，判断根节点是否满足条件，然后递归判断左右子节点是否满足条件。判断是否满足BST性质，可以从根节点下传节点的值域，root的初始值域是多少呢，(Integer.MIN_VALUE, Integer.MAX_VALUE)貌似是一个合理的选择，但如果root本身就是Integer.MIN_VALUE或者是Intger.MAX_VALUE呢？因此还需要使用两个辅助flag，effLowBound, effUpBound分别表示值域的上下边界是否有效，对于root，上下边界均无效。


public class Solution {
    public boolean isValidBST(TreeNode root) {
        return check(root, Integer.MIN_VALUE, Integer.MAX_VALUE, false, false);
    }
    public boolean check(TreeNode curr, int low, int up, 
            boolean effLowBound, boolean effUpBound) {
        if (curr == null) return true;
        if ((effLowBound && curr.val <= low) || (effUpBound && curr.val >= up))
            return false;
        return check(curr.left, low, curr.val, effLowBound, true)
                && check(curr.right, curr.val, up, true, effUpBound);
    }
}