Leetcode-437. Path Sum III

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

10
/  \
5   -3
/ \    \
3   2   11
/ \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

这个题目一开始理解错题意了，以为是从一个节点开始，左右都可以。正确的理解是从一个节点开始左右只能选一条路。目前选了一个比较蠢的方面，自底向上的找，其实自上而下是比较好的。Your runtime beats
4.33% of java submissions.

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private int count = 0;
public int pathSum(TreeNode root, int sum) {
help(root, sum);
return count;
}
private List<Integer> help(TreeNode root,int sum){

if(root == null) return new ArrayList<Integer>();

List<Integer> result = new ArrayList<Integer>();

List<Integer> rights = help(root.right,sum);
List<Integer> lefts = help(root.left,sum);
if(root.val == sum) count ++;
result.add(root.val);
int i = 0, j = 0;
for(int right : rights){
if(root.val + right == sum) count ++;
result.add(root.val + right);
}
for(int left : lefts){
if(root.val + left == sum) count ++;
result.add(root.val + left);
}

return result;
}
}