129. Sum Root to Leaf Numbers

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.

class Solution {
    public int sumNumbers(TreeNode root) {
        List<List<Integer>> res = new ArrayList();
        List<Integer> cur = new ArrayList();
        help(root, res, cur);
        int sum = 0;
        for(int i = 0; i < res.size(); i++){
            int curr = 0;
            for(int j = 0; j < res.get(i).size(); j++){
                curr = curr * 10 + res.get(i).get(j);
            }
            sum += curr;
        }
        return sum;
    }
    public void help(TreeNode root, List<List<Integer>> res, List<Integer> cur){
        if(root == null) return;
        cur.add(root.val);
        if(root.left == null && root.right == null){
            res.add(new ArrayList(cur));
        }
        help(root.left, res, cur);
        help(root.right, res, cur);
        cur.remove(cur.size() - 1);
    }
}

是113. path sumII的弱化版，本质是把所有遍历序列存到一个list（DFS），然后加在一起。cur是当前序列，每个数用完要remove掉防止重复。

public int sumNumbers(TreeNode root) {
    if (root == null) {
        return 0;
    }
    dfs(root, root.val);
    return sum;
}

int sum = 0;

private void dfs(TreeNode root, int cursum) {
    //到达叶子节点
    if (root.left == null && root.right == null) {
        sum += cursum;
        return;
    }
    //尝试左子树
    if(root.left!=null){
        dfs(root.left,  cursum * 10 + root.left.val);
    }
    //尝试右子树
    if(root.right!=null){
        dfs(root.right, cursum * 10 + root.right.val);

    }

}

另一种DFS，效率好像更高