501. Find Mode in Binary Search Tree

发表于 | 分类于

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).

解法1：In-order traversal

这题如果要求可以用extra space 的话很简单，用一个hashmap存储每一个出现的次数就可以了。
现在不能用extra space,那么根据bst的性质，一个中序遍历可以得到一个排序的数组。
对于这个排序数组，我们维护一个prev表示上一个节点，如果两个节点一样，我们更新count,如果不一样，则把当前的count和max相比，如果比max大则更新max并设置count为1.

C++

1

Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    TreeNode prev = null;
    int max = 1;
    int count = 1;
    public int[] findMode(TreeNode root) {
        int count = 0;
        TreeNode prev = null;
        ArrayList<Integer> res = new ArrayList<Integer>();
        traverse(root, res);
        int[] resArray = new int[res.size()];
        for (int i = 0; i < res.size(); i++) {
            resArray[i] = res.get(i);
        }
        return resArray;
    }
    private void traverse(TreeNode root, List<Integer> res) {
        if (root == null) {
            return;
        }
        traverse(root.left, res);
        if (prev != null) {
            if (root.val == prev.val) {
                count++;
            } else {
                count = 1;
            }
        }
        if (count > max) {
            res.clear();
            max = count;
            res.add(root.val);
        } else if (count == max) {
            res.add(root.val);
        }
        prev = root;
        traverse(root.right, res);        
    }
}