437. Path Sum III 二叉树路径的和3

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

  1. /**
  2. * Definition for a binary tree node.
  3. * public class TreeNode {
  4. * public int val;
  5. * public TreeNode left;
  6. * public TreeNode right;
  7. * public TreeNode(int x) { val = x; }
  8. * }
  9. */
  10. public class Solution {
  11. public int PathSum(TreeNode root, int sum) {
  12. if(root == null)
  13. return 0;
  14. return dfs(root, sum) + PathSum(root.left, sum) + PathSum(root.right, sum);
  15. }
  16. private int dfs(TreeNode root, int sum){
  17. int res = 0;
  18. if(root == null)
  19. return res;
  20. if(sum == root.val)
  21. res++;
  22. res+=dfs(root.left,sum - root.val);
  23. res+=dfs(root.right,sum - root.val);
  24. return res;
  25. }
  26. }







posted @ 2017-06-10 22:22  xiejunzhao  阅读(106)  评论(0编辑  收藏  举报