Path Sum III.java

E
1526529797
tags: Tree, DFS, Double Recursive

count所有存在的 path sum == target sum. 可以从任意点开始. 但是只能parent -> child .

#### DFS
- 对所给的input sum 做减法, 知道 sum 达到一个目标值截止
- 因为可以从任意点开始, 所以当sum达标时候, 需要继续recursive, 从而找到所有情况 (有正负数, sum可能继续增加/减少)
- 经典的 helper dfs recursive + self recursive
- 1. helper dfs recursive 处理包括root的情况
- 2. self recursive 来引领  skip root的情况.

#### 特点
- 与 `Binary Tree Longest Consecutive Sequence II` 在recursive的做法上很相似: 
- 利用dfs做包括root的recursive computation
- 利用这个function自己, 做`不包括root的recursive computation`

```
/*
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
*/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int pathSum(TreeNode root, int sum) {
        if (root == null) {
            return 0;
        }
        
        return dfs(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    }
    
    private int dfs(TreeNode node, int sum) {
        int count = 0;
        if (node == null) return count;

        count += node.val == sum ? 1 : 0;
        count += dfs(node.left, sum - node.val);
        count += dfs(node.right, sum - node.val);
        return count;
    }
}


```