Leetcode | Sum Root to Leaf Numbers

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.

这道题就很简单了。反正就是递归，每遍历到一个点，就把前面计算到的数*10+当前数，到叶子结点的时候就把数加一下。

叶子结点就是左右结点都为NULL。

 1 /**

 2  * Definition for binary tree

 3  * struct TreeNode {

 4  *     int val;

 5  *     TreeNode *left;

 6  *     TreeNode *right;

 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 8  * };

 9  */

10 class Solution {

11 public:

12     int sumNumbers(TreeNode *root) {

13         int sum = 0;   

14         recursive(root, 0, sum);

15         return sum;

16     }

17     

18     void recursive(TreeNode* root, int value, int& sum) {

19         if (root == NULL) {

20             return;

21         }

22         int v = value * 10 + root->val;

23         if (root->left == NULL && root->right == NULL) {

24             sum += v;

25         }

26         recursive(root->left, v, sum);

27         recursive(root->right, v, sum);

28     }

29 };

 第三遍刷，写了另外一种。

 1 /**

 2  * Definition for binary tree

 3  * struct TreeNode {

 4  *     int val;

 5  *     TreeNode *left;

 6  *     TreeNode *right;

 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 8  * };

 9  */

10 class Solution {

11 public:

12     int sumNumbers(TreeNode *root) {

13         return recurse(root, 0);

14     }

15     

16     int recurse(TreeNode *root, int sum) {

17         if (root == NULL) return 0;

18         if (root->left == NULL && root->right == NULL) return sum * 10 + root->val;

19         int left = recurse(root->left, sum * 10 + root->val);

20         int right = recurse(root->right, sum * 10 + root->val);

21         return left + right;

22     }

23 };