《leetCode》：Sum Root to Leaf Numbers

题目

Given a binary tree containing digits from - only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path ->-> which represents the number 

Find the total sum of all root-to-leaf numbers.

For example,

    
   / \
     
The root-to-leaf path -> represents the number 
The root-to-leaf path -> represents the number 

Return the sum =  +  = 
           

思路

第一步：找出所有的从root到叶子节点的路径

第二步：求和

实现代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private List<List<Integer>> res=new ArrayList<List<Integer>>();//用来保存所有的路径
    public int sumNumbers(TreeNode root) {
        if(root==null){
            return ;
        }
        //第一步：找出所有的路径
        sumNumbersHelper(root,new ArrayList<Integer>());
        //求和
        return sumAllList(res);
    }

    private int sumAllList(List<List<Integer>> listList) {
        if(listList==null){
            return ;
        }
        int sum=;
        int listSum=;
        for(List<Integer> temp:listList){
            listSum=;//注意：要在这里初始化为零
            for(Integer num:temp){
                listSum=listSum*+num;
            }
            sum+=listSum;
        }
        return sum;
    }

    private void sumNumbersHelper(TreeNode root,List<Integer> list) {
        if(root==null){
            return;
        }
        list.add(root.val);
        if(root.left==null&&root.right==null){//到达叶子节点，保存此条路径           
            res.add(list);
            return ;
        }
        else{
            sumNumbersHelper(root.left,new ArrayList<Integer>(list));
            sumNumbersHelper(root.right,new ArrayList<Integer>(list));
        }
    }
}
           

小结

原以为这道题自己做不出来的，写了写着思路就开了，太开兴了，哈哈