437. Path Sum III

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

思路：用分治的思想做，得到left的pathsum,right的pathsum，再加上当前节点的sum值，进行返回。当计算当前的sum值时，还需要一个递归：
/**
* 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 root, int sum) {
if(root == null){
return 0;
}
if(root.val == sum){
return 1 + dfs(root.left, sum - root.val) + dfs(root.right, sum - root.val);
}
return dfs(root.left, sum - root.val) + dfs(root.right, sum - root.val);
}
}