1 解題思想

這題我的做法，直接就是先序遍歷，遍歷過程中，檢查是否出錯了，程式碼很簡單。

注意用於上一個值追蹤的last，一開始必須要比Int的最小值小才行，所以用了long

2 原題

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.
Example 1:
    2
   / \
  1   3
Binary tree [2,1,3], return true.
Example 2:
    1
   / \
  2   3
Binary tree [1,2,3], return false.

3 AC解

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 

public class Solution {
    long last = Long.MIN_VALUE;
     //採用左中右遍歷後檢查是否遞增
    public boolean isValidBST(TreeNode root) {  
        if (root == null) return true;  
        if ( !isValidBST(root.left) ) return false;  
        if (root.val <= last) return false;
        last = root.val;
        return
 
 isValidBST(root.right) ;
    } 
}