Leetcode-129. Sum Root to Leaf Numbers

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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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.

这个题目一开始没想到可以传值的时候直接每次把上次的结果乘以10就好了。Your runtime beats 8.54% of java submissions.

public class Solution {    public int sumNumbers(TreeNode root) {        List<Integer> level = new ArrayList<Integer>();        return sumNumbers(root,level);    }    public int sumNumbers(TreeNode root, List<Integer> level){        int sum = 0;        if(root == null) return 0;        if(root.left == null && root.right == null){level.add(1);sum += root.val;}        else{            List<Integer> rightList = new ArrayList<Integer>();            List<Integer> leftList = new ArrayList<Integer>();            sum += sumNumbers(root.left,leftList);            sum += sumNumbers(root.right,rightList);            level.addAll(rightList);            level.addAll(leftList);            for(int i = 0; i < level.size(); i ++){                int item = level.get(i);                sum += (int)Math.pow(10,item) * root.val;                level.set(i,item + 1);            }        }        return sum;    }}

感觉自己对于数据结构的本质理解还是不够深。Your runtime beats 61.60% of java submissions.

public class Solution {    public int sumNumbers(TreeNode root) {return sumNumbers(root,0);            }    public int sumNumbers(TreeNode root, int lastResults){if(root == null) return 0;if(root.left == null && root.right == null) return lastResults * 10 + root.val;return sumNumbers(root.left,lastResults * 10 + root.val) + sumNumbers(root.right,lastResults * 10 + root.val);    }}