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).
题目标签：Tree
这道题目给了我们一个二叉搜索树，让我们找到树的众树 （出现最多的那个值），可以是一个众树，也可以有很多个。看完题目第一个想到用HashMap，但是发现题目最后follow up说不能用extra space。所以我们要另外考虑方法。二叉搜索树的特性，左边 <= 根 <= 右边，这道题目包括了等于。举一个例子：
10
/ \
5 13
/ \ \
3 7 13
/ \ \
2 3 9
看这个例子，我们试着把它上下压缩一下， 就等于， 2 3 3 5 7 9 10 13 13 ，在纸上画的左右分开比较容易看清。发现这是一个有序的排列。如果我们可以遍历这个顺序的话，它是从小到大的，特点就是，一样的数字一定是连在一起的。这样我们就可以设一个count = 1，一个maxCount = 0 和一个pre Node， count是记录每一个数字的连续出现次数，如果大于maxCount 那就说明这个数字是新的mode，比起之前的数字。 maxCount 记录最大出现次数的mode。pre Node是上一个点的数字，当每次current 点和上一个点比较，是否两个数字相同，来判断需要count++，如果不相同，那就更新count = 1。
如何得到这个有序排列，可以通过inorder traversal 来实现，需要注意的是， Java 是 Pass by Value 的核心，所以pre node ， count 什么的，需要放在function 外面。
Java Solution:
Runtime beats 74.31%
完成日期：07/07/2017
关键词：Tree
关键点：inorder 遍历 （从小到大顺序）
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 }
参考资料：
http://www.cnblogs.com/grandyang/p/6436150.html
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