508. Most Frequent Subtree Sum 出现次数最多的子树元素和

@TOC

# 题目描述

Given the root of a tree, you are asked to find the most frequent subtree sum. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.

Examples 1

``````Input:

5
/  \
2   -3
return [2, -3, 4], since all the values happen only once, return all of them in any order.
Examples 2
Input:

5
/  \
2   -5
return [2], since 2 happens twice, however -5 only occur once.
``````

Note: You may assume the sum of values in any subtree is in the range of 32-bit signed integer.

# 解题方法

Python解法如下：

``````# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def findFrequentTreeSum(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if not root: return []
vals = []
def getSum(root):
if not root:
return 0
s = getSum(root.left) + root.val + getSum(root.right)
vals.append(s)
# 别忘了返回s
return s
getSum(root)
count = collections.Counter(vals)
frequent = max(count.values())
return [x for x, v in count.items() if v == frequent]
``````

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> findFrequentTreeSum(TreeNode* root) {
dfs(root);
map<int, int> d;
int most_common = 0;
for (int s : sums) {
d[s] ++;
most_common = max(most_common, d[s]);
}
vector<int> res;
for (auto p : d){
if (p.second == most_common)
res.push_back(p.first);
}
return res;
}
private:
vector<int> sums;
void dfs(TreeNode* root) {
if (!root) return;
dfs(root->left);
sums.push_back(getSum(root));
dfs(root->right);
}
int getSum(TreeNode* root) {
if (!root) return 0;
int l = getSum(root->left);
int r = getSum(root->right);
return root->val + l + r;
}
};
``````

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> findFrequentTreeSum(TreeNode* root) {
getSum(root);
vector<int> res;
for (auto p : d)
if (p.second == most_common)
res.push_back(p.first);
return res;
}
private:
vector<int> sums;
int most_common = 0;
map<int, int> d;
int getSum(TreeNode* root) {
if (!root) return 0;
int l = getSum(root->left);
int r = getSum(root->right);
int s = root->val + l + r;
sums.push_back(s);
d[s]++;
most_common = max(most_common, d[s]);
return s;
}
};
``````

# 日期

2018 年 3 月 4 日 2018 年 12 月 13 日 —— 时间匆匆，如何才能提高时间利用率？