题目
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.
找众数
用use HashMap<Integer, frequency>
(List??)
返回int[]
traverse: inorder, pre, post - anyone is fine
O(n)时间复杂度
R:
Java:Map与HashMap,Hashtable,HashSet比较
- List in Java
有序、可重复; 可以持续扩展大小(array不可以)
ListList集合代表一个元素有序、可重复的集合,集合中每个元素都有其对应的顺序索引。注意: List集合默认按元素的添加顺序设置元素索引,例如第 ...
==> 用List的原因是:List可以持续扩展大小,而且数组不可以。
Java数组声明后其长度就固定了,不能增加长度。 要创建一个可扩展的数组可以使用ArrayList或Vector。
class Solution {
HashMap<Integer, Integer> map;
// = new HashMap<>();
int max = 0;
public int[] findMode(TreeNode root) {
if(root == null) {return new int[0];}
this.map = new HashMap<>();
inorder(root);
List<Integer> list = new LinkedList<>();
for(int key: map.keySet()) {
if(map.get(key) == max) {list.add(key);}
}
int[] res = new int[list.size()];
for(int i = 0; i<res.length; i++) res[i] = list.get(i);
// transfer the LinkedList list to int[] res
return res;
}
private void inorder(TreeNode node) {
if(node.left != null) {inorder(node.left);}
map.put(node.val, map.getOrDefault(node.val, 0) + 1);
max = Math.max(max, map.get(node.val));
// constantly update the maximum node.val
if(node.right != null) {inorder(node.right);}
}
}
评价:
O(n) Time, O(n) Space;
it doesn't use the property of BST.
// P⊆NP⊆PSPACE=NPSPACE⊆EXPTIME
O(1) Space
public class Solution {
public int[] findMode(TreeNode root) {
inorder(root);
modes = new int[modeCount];
modeCount = 0;
currCount = 0;
inorder(root);
return modes;
}
private int currVal;
private int currCount = 0;
private int maxCount = 0;
private int modeCount = 0;
private int[] modes;
private void handleValue(int val) {
if (val != currVal) {
currVal = val;
currCount = 0;
}
currCount++;
if (currCount > maxCount) {
maxCount = currCount;
modeCount = 1; // BSTinorder遍历时,单调递增的属性
} else if (currCount == maxCount) {
if (modes != null)
modes[modeCount] = currVal;
modeCount++;
}
}
private void inorder(TreeNode root) {
if (root == null) return;
inorder(root.left);
handleValue(root.val);
inorder(root.right);
}
}
遍历了两遍,但是只用了一个modes[],只占用O(1)Space。
第一遍找到众数with highest frequency,并且记为N;
第二遍将所有出现了N次的数组都存在一个modes里。
I think the way to do it properly is to do two passes.
- One to find the highest number of occurrences of any value, and then a
- second pass to collect all values occurring that often.
https://leetcode.com/problems/find-mode-in-binary-search-tree/discuss/98101/Proper-O(1)-space
每次遍历一个数字Val时:
currVal = Val, 记录当前数字出现的次数currCount;
对得到对currCount进行判断:
是否出现次数大于maxCount,是则 - 1. 更新最大频率maxCount,2. 新的maxCount 对应的数字个数 更新为1(后期出现频率相同的数字时modeCount++)。