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.
Note: A leaf is a node with no children.
Example:
Input: [1,2,3] 1 / \ 2 3 Output: 25 Explanation: The root-to-leaf path 1->2 represents the number 12. The root-to-leaf path 1->3 represents the number 13. Therefore, sum = 12 + 13 = 25.Example 2:
Input: [4,9,0,5,1] 4 / \ 9 0 / \ 5 1 Output: 1026 Explanation: The root-to-leaf path 4->9->5 represents the number 495. The root-to-leaf path 4->9->1 represents the number 491. The root-to-leaf path 4->0 represents the number 40. Therefore, sum = 495 + 491 + 40 = 1026.前面正好做了257打印出所有路径，这道题就只是把这些打印出来的路径变成int value然后相加就行了。直接这么做了先。
Runtime: 1 ms, faster than 29.35% of Java online submissions for Sum Root to Leaf Numbers.
Memory Usage: 36.6 MB, less than 16.49% of Java online submissions for Sum Root to Leaf Numbers.
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public int sumNumbers(TreeNode root) { List<String> paths = binaryTreePaths(root); int result = 0; for (String s : paths) { result += Integer.valueOf(s); } return result; } private List<String> binaryTreePaths(TreeNode root) { List<String> result = new ArrayList<>(); if (root == null) { return result; } helper(root, new ArrayList<>(), result); return result; } public void helper(TreeNode root, List<Integer> temp, List<String> result) { temp.add(root.val); if (root.left == null && root.right == null) { StringBuilder sb = new StringBuilder(); for (int i : temp) { sb.append(i); } result.add(sb.toString()); return; } if (root.left != null) { helper(root.left, temp, result); temp.remove(temp.size() - 1); } if (root.right != null) { helper(root.right, temp, result); temp.remove(temp.size() - 1); } } }然后自己又重新写了个递归版本的，不需要像上面那个那样add又remove了（盲猜是因为之前的是pass by reference所以temp会被更新，这里是pass by value所以不用管？），但是更新result要把result写成global variable才能真正被更新到。
Runtime: 0 ms, faster than 100.00% of Java online submissions for Sum Root to Leaf Numbers.
Memory Usage: 36.5 MB, less than 16.49% of Java online submissions for Sum Root to Leaf Numbers.
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { int result = 0; public int sumNumbers(TreeNode root) { if (root == null) { return result; } helper(root, 0); return result; } private void helper(TreeNode root, int temp) { temp = temp * 10 + root.val; if (root.left == null && root.right == null) { result += temp; } if (root.left != null) { helper(root.left, temp); } if (root.right != null) { helper(root.right, temp); } } }看了discussion，其实还可以直接让helper return int，想起来稍微有点费劲，但写起来是真的优雅。
Runtime: 0 ms, faster than 100.00% of Java online submissions for Sum Root to Leaf Numbers.
Memory Usage: 36.9 MB, less than 16.49% of Java online submissions for Sum Root to Leaf Numbers.
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public int sumNumbers(TreeNode root) { return helper(root, 0); } private int helper(TreeNode root, int temp) { if (root == null) { return 0; } temp = temp * 10 + root.val; if (root.left == null && root.right == null) { return temp; } return helper(root.left, temp) + helper(root.right, temp); } }
