Question:

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

  1. The left subtree of a node contains only nodes with keys less than the node’s key.
  2. The right subtree of a node contains only nodes with keys greater than the node’s key.
  3. Both the left and right subtrees must also be binary search trees.
Example 1:

Input:
    2
   / \
  1   3
Output: true

Example 2:

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

Thinking:

  • Method:
    • 递归遍历，要确定是否为valid则要确定当前节点的值是否在范围之间。
      • 范围初始化值为Long.MAX_VALUE和Long.MIN_VALUE
      • 对于左子树而言，上限就是当前的结点。
      • 对于右子树而言，下限就是当前的结点。
      • 遍历到null（叶子结点），则说明为true，这也是退出递归的条件。
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        return check(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    private static boolean check(TreeNode node, long min, long max){
        if(node == null) return true;
        if(node.val >= max || node.val <= min) return false;
        return check(node.left, min, node.val) && check(node.right, node.val, max);
    }
}

二刷

这道题写出来还是很快的，对于这种树的判断，大多是使用递归的方法。但是初始的最大值和最小值我还是用Integer类型的，所以会有Error。以后要将这样的情况考虑进来。

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