标签：dup dia only any def ++ integer als set
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:
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).
M1: using extra space
traverse的时候，用hashmap存每个节点出现的次数
time: O(n), space: O(n)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { int max = 0; public int[] findMode(TreeNode root) { if(root == null) { return new int[] {}; } Map<Integer, Integer> map = new HashMap<>(); traverse(root, map); List<Integer> tmp = new ArrayList<>(); for(Map.Entry<Integer, Integer> entry : map.entrySet()) { if(entry.getValue() == max) { tmp.add(entry.getKey()); } } int[] res = new int[tmp.size()]; for(int i = 0; i < tmp.size(); i++) { res[i] = tmp.get(i); } return res; } public void traverse(TreeNode root, Map<Integer, Integer> map) { if(root == null) { return; } map.put(root.val, map.getOrDefault(root.val, 0) + 1); max = Math.max(max, map.get(root.val)); traverse(root.left, map); traverse(root.right, map); } }
M2: optimized, not using extra space
501. Find Mode in Binary Search Tree - Easy
标签：dup dia only any def ++ integer als set
原文地址：https://www.cnblogs.com/fatttcat/p/10200168.html
