LeetCode 501. Find Mode in Binary Search Tree （找到二叉搜索树的众数）

 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 public class Solution 
11 {
12     TreeNode pre = null;
13     int cnt = 1;
14     int max_cnt = 0;
15     
16     public int[] findMode(TreeNode root) 
17     {
18         ArrayList<Integer> res = new ArrayList<>();
19         
20         inorder(root, res);
21         
22         int[] result = new int[res.size()];
23         
24         for(int i=0; i<result.length; i++)
25             result[i] = res.get(i);
26         
27         return result;
28     }
29     
30     public void inorder(TreeNode node, ArrayList<Integer> res)
31     {
32         if(node == null)
33             return;
34         
35         inorder(node.left, res);
36         // meaning this node has a previous node, need to compare them to determine cnt
37         if(pre != null) 
38         {    
39             if(node.val == pre.val) // if this node has same value as pre‘s
40                 cnt++;
41             else // if this node has different value as pre‘s
42                 cnt = 1;
43         }
44         
45         // once cnt is greater max_cnt, meaning find a new mode, need to clear res;
46         if(cnt > max_cnt) 
47         {
48             max_cnt = cnt; 
49             res.clear();
50             res.add(node.val);
51         }
52         else if(cnt == max_cnt) // cnt == max_cnt, meaning find a new mode that equal to pre mode.
53             res.add(node.val);
54         
55             
56         if(pre == null) // for first most left leaf node, its pre is null, set the first pre node 
57         {    
58             pre = new TreeNode(node.val);
59         }
60         else // if pre is not null, update this node‘s val each time
61             pre.val = node.val;
62         
63         
64         inorder(node.right, res);
65     }
66 }