【LeetCode】129. Sum Root to Leaf Numbers_sum of numbers from 0 to n-优快云博客

本文探讨了如何在仅包含0-9数字的二叉树中寻找所有从根到叶的路径所代表的数字总和。通过递归方法，文章详细解释了如何计算这些路径的数值，并提供了一个C++实现的代码示例。

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.

题解：回溯的简单题，设置一个tmp表示递归到叶子节点中存储的路径数，sum表示和，判断叶子节点后变量tmp并加入到sum中，然后递归出来后把tmp弹出来

代码：

class Solution {
public:
    int sumNumbers(TreeNode* root) {
        int sum=0;
        vector<int> tmp;
        dfs(root,sum,tmp);
        return sum;
    }
    void dfs(TreeNode* root,int& sum,vector<int>& tmp)
    {
        if(root==NULL) return;
        tmp.push_back(root->val);
        if(root->left==NULL&&root->right==NULL){
            int ans=0;int j=1;
            for(int i=0;i<tmp.size();i++){
                ans+=tmp[i];
                if(i!=tmp.size()-1) ans*=10;
            }
            sum+=ans;
        }
        dfs(root->left,sum,tmp);
        dfs(root->right,sum,tmp);
        tmp.pop_back();
    }
};