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lxzgitcode
Leetcode-Java
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master
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Leetcode-Java
/
leetcode
/
solution
/
src
/
FindModeInBinarySearchTree.java
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65 lines (50 loc) · 1.8 KB
master
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Leetcode-Java
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leetcode
/
solution
/
src
/
FindModeInBinarySearchTree.java
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import
java
.
util
.
LinkedList
;
import
java
.
util
.
List
;
/**
* 这题是要找BST中出现次数最多的节点集合,这里允许有重复节点
* 思路很简单,中序遍历,会按升序排列,再统计重复的值
* 但是这样会消耗额外空间,对于类似于1,2,3,4...n-1, n-1,这种,所有的元素都只出现1次,
* 唯独n-1出现了两次,这样mList中要保存整棵树的元素,所以空间是O(n),而返回值事实上只有一个
* 这是分配在堆上的而非栈上,不能忽略。
*
* 如果额外空间要求为O(l),则分开两次扫描,第一次扫描最大次数,不记录最大次数的元素集合
* 第二次根据最大次数来收集元素
*/
public
class
FindModeInBinarySearchTree
{
/**
* 这题类似于在一个有序的序列中查找频率最高的元素集合
*/
private
List
<
Integer
>
mList
;
private
int
mCurCount
;
private
int
mMaxCount
;
private
Integer
mCurValue
;
public
int
[]
findMode
(
TreeNode
root
) {
mList
=
new
LinkedList
<>();
inorderTraverse
(
root
);
int
[]
res
=
new
int
[
mList
.
size
()];
for
(
int
i
=
0
;
i
<
res
.
length
;
i
++) {
res
[
i
] =
mList
.
get
(
i
);
}
return
res
;
}
private
void
inorderTraverse
(
TreeNode
root
) {
if
(
root
==
null
) {
return
;
}
inorderTraverse
(
root
.
left
);
if
(
mCurValue
!=
null
&&
root
.
val
!=
mCurValue
) {
mCurCount
=
1
;
}
else
{
mCurCount
++;
}
mCurValue
=
root
.
val
;
if
(
mCurCount
>
mMaxCount
) {
mList
.
clear
();
mList
.
add
(
mCurValue
);
mMaxCount
=
mCurCount
;
}
else
if
(
mCurCount
==
mMaxCount
) {
mList
.
add
(
mCurValue
);
}
inorderTraverse
(
root
.
right
);
}
}
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