Sum Root to Leaf Numbers

Sum Root to Leaf Numbers

问题：

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.

思路：

　　dfs

我的代码：

public class Solution {
    public int sumNumbers(TreeNode root) {
        if(root == null) return 0;
        List<String> list = new ArrayList<String>();
        dfs(list, "", root);
        int sum = 0;
        for(String s : list)
        {
            sum += Integer.valueOf(s);
        }
        return sum;
    }
    public void dfs(List<String> list, String s, TreeNode root)
    {
        if(root == null)
            return;
        if(root.left == null && root.right == null)
        {
            list.add(s+root.val);
            return;
        }
        int val = root.val;
        dfs(list, s + val, root.left);
        dfs(list, s + val, root.right);
    }
}

他人代码：

public class Solution {
    public int sumNumbers(TreeNode root) {
        return dfs(root, 0);
    }

    private int dfs(TreeNode root, int prev){
        if(root == null) {
            return 0;
        }

        int sum = root.val + prev * 10;
        if(root.left == null && root.right == null) {
            return sum;
        }

        return dfs(root.left, sum) + dfs(root.right, sum);
    }
}

学习之处：

• 他人代码里面通过记录prev，进行数值的串联，一来省去了List的存储空间，二来不用Integer To String了