LeetCode -- Path Sum III
题目描述：
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.
给定一个二叉树，遍历过程中收集所有可能路径的和，找出和等于X的路径树。
思路：
设当前节点为root,分别收集左右节点路径和的集合，merge到当前集合中；
将当前节点添加到数组中，构成新的可能路径。
实现代码：
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.
给定一个二叉树，遍历过程中收集所有可能路径的和，找出和等于X的路径树。
思路：
设当前节点为root,分别收集左右节点路径和的集合，merge到当前集合中；
将当前节点添加到数组中，构成新的可能路径。
实现代码：
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private int _sum;
private int _count;
public int PathSum(TreeNode root, int sum)
{
_count = 0;
_sum = sum;
Travel(root, new List<int>());
return _count;
}
private void Travel(TreeNode current, List<int> ret){
if(current == null){
return ;
}
if(current.val == _sum){
_count ++;
}
var left = new List<int>();
Travel(current.left, left);
var right = new List<int>();
Travel(current.right, right);
ret.AddRange(left);
ret.AddRange(right);
for(var i = 0;i < ret.Count; i++){
ret[i] += current.val;
if(ret[i] == _sum){
_count ++;
}
}
ret.Add(current.val);
//Console.WriteLine(ret);
}
}