题目：
Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3 which represents the number 123.
Find the total sum of all root-to-leaf numbers.
For example,
1 / 2 3
The root-to-leaf path 1->2 represents the number 12. The root-to-leaf path 1->3 represents the number 13.
Return the sum = 12 + 13 = 25.
链接： http://leetcode.com/problems/sum-root-to-leaf-numbers/
6/8/2017
3ms, 9%
很长时间想不起来怎么做，把tree想成graph就搞定了
注意
1. 第22行还是要加上叶结点的值
2. 第25，32行需要加上当前节点的值，然后删掉。
1 public class Solution { 2 public int sumNumbers(TreeNode root) { 3 if (root == null) { 4 return 0; 5 } 6 List<Integer> ret = new ArrayList<Integer>(); 7 List<Integer> list = new ArrayList<Integer>(); 8 dfs(root, ret, list); 9 int sum = 0; 10 for (int i = 0; i < ret.size(); i++) { 11 sum += ret.get(i); 12 } 13 return sum; 14 } 15 private void dfs(TreeNode root, List<Integer> ret, List<Integer> list) { 16 if (root.left == null && root.right == null) { 17 int number = 0; 18 for (int i = 0; i < list.size(); i++) { 19 number = number * 10 + list.get(i); 20 } 21 number = number * 10 + root.val; 22 ret.add(number); 23 return; 24 } 25 list.add(root.val); 26 if (root.left != null) { 27 dfs(root.left, ret, list); 28 } 29 if (root.right != null) { 30 dfs(root.right, ret, list); 31 } 32 list.remove(list.size() - 1); 33 } 34 }
别人更简单的做法：
https://discuss.leetcode.com/topic/6731/short-java-solution-recursion
1 public int sumNumbers(TreeNode root) { 2 return sum(root, 0); 3 } 4 5 public int sum(TreeNode n, int s){ 6 if (n == null) return 0; 7 if (n.right == null && n.left == null) return s*10 + n.val; 8 return sum(n.left, s*10 + n.val) + sum(n.right, s*10 + n.val); 9 }
更多讨论
https://discuss.leetcode.com/category/137/sum-root-to-leaf-numbers