LeetCode-98. Validate Binary Search Tree

题目描述

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.

解题思路

判定一棵树是不是二叉查找树，个人感觉应该自下而上的判断，因为判断root节点是否满足条件时，不只是满足root.left.val小于root.val，和root.right.val大于root.val关系，而是还要比较左子树的最大值要小于root.val，同时右子树的最小值要大于root.val。本人自定义了数据类，自下而上的返回各个子树中的最大值和最小值。

代码

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
int max = Integer.MIN_VALUE;
int min = Integer.MAX_VALUE;
public boolean isValidBST(TreeNode root) {
if(root == null)
return true;
return validCore(root).flag;
}
public Result validCore(TreeNode root){
Result left = null;
if(root.left == null){
left = new Result();
left.min = root.val;
left.max = root.val;
}else{
left = validCore(root.left);
if(!left.flag || left.max >= root.val){
left.flag = false;
return left;
}
}
Result right = null;
if(root.right == null){
right = new Result();
right.min = root.val;
right.max = root.val;
}else{
right = validCore(root.right);
if(!right.flag || right.min <= root.val){
right.flag = false;
return right;
}
}
right.max = Math.max(right.max,root.val);
right.min = Math.min(left.min,root.val);
return right;
}
}
class Result{
boolean flag = true;
int max = Integer.MIN_VALUE;
int min = Integer.MAX_VALUE;
}