题目：
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
链接：https://leetcode.com/problems/path-sum-iii/#/description
3/25/2017
这道题自己想出来了思路，然而不会写后序遍历。。。
出现的问题：
1. 不会写iterative的后序遍历：（
2. Vector不好用，以后刷题也不用了。
3. 对几种主要的Collections的方法不熟悉
思路：后序遍历iterative时stack含有从根节点到当前节点的路径。我们用arraylist取代stack，每当有新元素加到stack时从尾到头看和是否为target，是的话＋1。一个stack可能有多个符合要求的路径（比如某节点值为0，或几个节点和为0）
performance 23ms，但是我认为这个应该算是中等题，简单题有点为难我了。
时间复杂度O(nlgn)，worst caseO(n^2)
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public int pathSum(TreeNode root, int sum) { 12 ArrayList<TreeNode> v = new ArrayList<TreeNode>(); 13 if (root == null) return 0; 14 TreeNode prev = null; // previously traversed node 15 TreeNode curr = root; 16 int count =0; 17 18 v.add(root); 19 if (root.val == sum) count++; 20 21 while (!v.isEmpty()) { 22 curr = v.get(v.size()-1); 23 if (prev == null || prev.left == curr || prev.right == curr) { // traverse down the tree 24 if (curr.left != null) { 25 v.add(curr.left); 26 count += getCurrentCount(v, sum); 27 } else if (curr.right != null) { 28 v.add(curr.right); 29 count += getCurrentCount(v, sum); 30 } 31 } else if (curr.left == prev) { // traverse up the tree from the left 32 if (curr.right != null) { 33 v.add(curr.right); 34 count += getCurrentCount(v, sum); 35 } 36 } else { // traverse up the tree from the right 37 v.remove(v.size()-1); 38 } 39 prev = curr; 40 } 41 return count; 42 } 43 44 private int getCurrentCount(ArrayList<TreeNode> v, int sum) { 45 int count = 0; 46 int tempSum = 0; 47 for (int i = v.size() - 1; i >= 0; i--) { 48 tempSum += v.get(i).val; 49 if (tempSum == sum) count++; 50 } 51 return count; 52 } 53 }
别人用hashmap储存prefix sum，我还没有看懂：
https://discuss.leetcode.com/topic/64526/17-ms-o-n-java-prefix-sum-method/11
更简单的recursive写法：
https://discuss.leetcode.com/topic/64388/simple-ac-java-solution-dfs
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