129. Sum Root to Leaf Numbers

problem description

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.

algorithm thought

树问题，递归解决，这里需要到叶子节点才能知道最后的值是多少。需要将父节点的信息传入子节点。这里有两种方法传递一种是直接改变节点的值，父节点改将自己的值加到子节点。还有一种方法是用参数传递。

code

//改变叶子节点的值
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int res=0;
    int sumNumbers(TreeNode* root) {        
        if(root!=NULL)
            sumNumber(root);
        return res;
    }
    void sumNumber(TreeNode* root){
        if(root->left){
            root->left->val+=root->val*10;
            sumNumber(root->left);
        }
        if(root->right){
            root->right->val+=root->val*10;
            sumNumber(root->right);
        }
        if(root->left==NULL&&root->right==NULL)
            res+=root->val;
    }
};

//用参数传递
class Solution {
public: 
    int res=0;
    int sumNumbers(TreeNode* root) {        
        if(root==NULL)
            return 0;
        helper(root,0);
        return res;
    }
    void helper(TreeNode* root,int add){
        if(root->left==NULL&&root->right==NULL){
            res+=(add*10+root->val);
            return;
        }
        if(root->left)
            helper(root->left,add*10+root->val);
        if(root->right)
            helper(root->right,add*10+root->val);
    }
};

algorithm analysis

两个代码时间复杂度分析，可能大家都认为第一个算法更好。但是运行下来，反而是第二个代码更快，我认为是第一个代码修改了指针中的值，修改值可能比读取一个值消耗的时间多很多，导致用第二个代码传参数更好。