LeetCode 98. Validate Binary Search Tree（校验二叉搜索树）

原题网址：https://leetcode.com/problems/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.

方法一：递归+自顶向下。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private boolean isValid = true;
    private void valid(TreeNode root, Check check) {
        if (!isValid) return;
        if (root == null) return;
        check.min = root.val;
        check.max = root.val;
        if (root.left != null) {
            Check left = new Check();
            valid(root.left, left);
            check.min = left.min;
            if (left.max >= root.val) {
                isValid = false;
                return;
            }
        }
        if (!isValid) return;
        if (root.right != null) {
            Check right = new Check();
            valid(root.right, right);
            check.max = right.max;
            if (right.min <= root.val) {
                isValid = false;
                return;
            }
        }
    }
    public boolean isValidBST(TreeNode root) {
        valid(root, new Check());
        return isValid;
    }
}
class Check {
    int min;
    int max;
}

方法二：递归+分治策略

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private boolean valid(TreeNode root, int min, int max) {
        if (root == null) return true;
        if (root.val < min || root.val > max) return false;
        return 
        ((root.val > Integer.MIN_VALUE && valid(root.left, min, root.val-1))
        || (root.val == Integer.MIN_VALUE && root.left == null))
        && 
        ((root.val < Integer.MAX_VALUE && valid(root.right, root.val+1, max))
        || (root.val == Integer.MAX_VALUE && root.right == null));
    }
    public boolean isValidBST(TreeNode root) {
        return valid(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
    }
}

方法三：按照遍历顺序检查。

/**
 * 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 = null;
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        if (root.left != null) {
            boolean left = isValidBST(root.left);
            if (!left) return false;
        }
        if (prev != null && prev.val >= root.val) return false;
        prev = root;
        if (root.right != null) {
            boolean right = isValidBST(root.right);
            if (!right) return false;
        }
        return true;
    }
}