129. Sum Root to Leaf Numbers

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.

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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。

public class Solution {
    public int sumNumbers(TreeNode root) {
        if (root == null) {
            return 0;
        }
        List<Integer> ret = new ArrayList<Integer>();
        List<Integer> list = new ArrayList<Integer>();
        dfs(root, ret, list);
        int sum = 0;
        for (int i = 0; i < ret.size(); i++) {
            sum += ret.get(i);
        }
        return sum;
    }
    private void dfs(TreeNode root, List<Integer> ret, List<Integer> list) {
        if (root.left == null && root.right == null) {
            int number = 0;
            for (int i = 0; i < list.size(); i++) {
                number = number * 10 + list.get(i);
            }
            number = number * 10 + root.val;
            ret.add(number);
            return;
        }
        list.add(root.val);
        if (root.left != null) {
            dfs(root.left, ret, list);
        }
        if (root.right != null) {
            dfs(root.right, ret, list);
        }
        list.remove(list.size() - 1);
    }
}

别人更简单的做法：

https://discuss.leetcode.com/topic/6731/short-java-solution-recursion

public int sumNumbers(TreeNode root) {
    return sum(root, 0);
}

public int sum(TreeNode n, int s){
    if (n == null) return 0;
    if (n.right == null && n.left == null) return s*10 + n.val;
    return sum(n.left, s*10 + n.val) + sum(n.right, s*10 + n.val);
}

更多讨论

https://discuss.leetcode.com/category/137/sum-root-to-leaf-numbers