501. Find Mode in Binary Search Tree【LeetCode by java】

Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST.

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than or equal to the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

 

For example:
Given BST [1,null,2,2],

   1
    \
     2
    /
   2

 

return [2].

Note: If a tree has more than one mode, you can return them in any order.

Follow up: Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).

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解题：给一个二叉查找树，返回出现次数最多的元素（众数）。并且要求不使用额外的空间，要求结果以数组的形式返回。我使用了一个ArrayList，有额外的空间消耗，虽然不满足条件，但是比较简单方便。当然也可以把数组定义为成员变量，由于数组的长度不知道，还得定义一个末尾指针，太麻烦，不如用集合。思路是，遍历BST，由于其本身的性质，遍历的过程中得到的序列，就是一个有序序列。定一个max用来保存出现做多的次数，pre记录父节点，如果没有，则为空。再定义一个res的集合保存结果，定一个cns保存当前出现最多的次数（暂时），然后不断地比较cns和max的值，随时更新res结果集，代码如下：

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     ArrayList<Integer>res = null;
12     TreeNode pre = null;
13     int max = 0;
14     int cns = 1;
15     public int[] findMode(TreeNode root) {
16         //存放结果
17         res = new ArrayList<Integer>();
18         middle_search(root);
19         int[] arr =new int[res.size()]; 
20         for(int i =0; i < arr.length; i++)
21             arr[i] = res.get(i);
22         return arr;
23     }
24     public void middle_search(TreeNode root) {
25         if(root == null)
26             return ;
27         middle_search(root.left);    
28         //处理根结点
29         if(pre != null){  //有父节点
30             if(root.val == pre.val)
31                 cns++;
32             else cns = 1;
33         }
34         if(cns >= max){
35             if(cns > max)
36                res.clear();
37             max = cns;
38             res.add(root.val);
39         }
40         pre = root;
41         
42         middle_search(root.right);
43     }
44 }