Medium
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.
太久没有尝过不看答案直接做出来的滋味了，这道题满足了一下我。看了题之后很容易就联想到了pathSum这一类跟root to leaf path相关的题，所以思路也比较清晰了，就是用DFS把所有路径遍历出来，最后来处理数字和加法就可以了。 这里依旧要注意一下,res.add(path)是不行的，因为path变空了之后res里面将什么也没有, 必须要写成res.add(new ArrayList<>(path)). 这样不管path之后怎么变，我copy过来的path永远会保持现在的样子，相当于我只是把path里的元素copy过来了，并没有copy path的reference。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int sumNumbers(TreeNode root) {
int sum = 0;
List<Integer> path = new ArrayList<>();
List<List<Integer>> res = new ArrayList<>();
if (root == null){
return sum;
}
helper(root, path, res);
for (List<Integer> list : res){
int pathSum = 0;
for (Integer i : list){
pathSum *= 10;
pathSum += i;
}
sum += pathSum;
}
return sum;
}
private void helper(TreeNode root, List<Integer> path, List<List<Integer>> res){
if (root == null){
return;
}
path.add(root.val);
if (root.left == null && root.right == null){
res.add(new ArrayList<>(path));
}
helper(root.left, path, res);
helper(root.right, path, res);
path.remove(path.size() - 1);
}
}
