【原题】 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.
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.
【解释】 给一颗二叉树,要求返回其从根结点到叶子结点组成数字之和。从根结点到叶子结点从上到下对应从左到右。例如路径1->2,则代表数字为12. 【思路】 基本上和Path Sum II同样的思路,上篇博文也有。思路基本相同,只不过在找到满足条件的子集之后,本题要做的是加和。
public class Solution {
private int sum=0;
public int getNum(List<Integer> list){//得到所在list所有值的正数表示
int res=0;
for(int i=0;i<list.size();i++){
res=res*10+list.get(i);
}
return res;
}
public void sumNumbersCore(List<Integer> list, TreeNode root){
list.add(root.val);
if(root.left==null&&root.right==null)
sum+=getNum(list);//将所有路径代表的数字累加
else{
if(root.left!=null)
sumNumbersCore(list,root.left);
if(root.right!=null)
sumNumbersCore(list,root.right);
}
list.remove(list.size()-1);
}
public int sumNumbers(TreeNode root) {
if(root==null) return 0;//空树
List<Integer> list=new ArrayList<Integer>();
sumNumbersCore(list,root);
return sum;
}
}