LeetCode - Sum Root to Leaf Numbers

Sum Root to Leaf Numbers

2014.1.1 19:55

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.

Solution:

　　The solution is plain, traverse the tree and add up the numbers at leaf nodes.

　　1->2->3 forms "123", that's (1 * 10 + 2) * 10 + 3. Then you know how the recursion is done.

　　Time and space complexities are both O(n), where n is the number of nodes in the tree. The space complexity comes from the local paramater in recursive calls.

Accepted code:

 1 //1CE, 2WA, 1AC

 2 /**

 3  * Definition for binary tree

 4  * struct TreeNode {

 5  *     int val;

 6  *     TreeNode *left;

 7  *     TreeNode *right;

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

 9  * };

10  */

11 class Solution {

12 public:

13     int sumNumbers(TreeNode *root) {

14         // IMPORTANT: Please reset any member data you declared, as

15         // the same Solution instance will be reused for each test case.

16         

17         if(root == nullptr){

18             return 0;

19         }

20         

21         result = 0;

22         traverse(root, root->val);

23         

24         return result;

25     }

26 private:

27     int result;

28     void traverse(TreeNode *root, int weight) {

29         // Which level should 'weight' represent, vague...

30         // That's why you got two WAs here!!!

31         if(root == nullptr){

32             return;

33         }

34         

35         if(root->left == nullptr && root->right == nullptr){

36             result += weight;

37             return;

38         }

39         

40         if(root->left != nullptr){

41             traverse(root->left, weight * 10 + root->left->val);

42         }

43         if(root->right != nullptr){

44             traverse(root->right, weight * 10 + root->right->val);

45         }

46     }

47 };