Leetcode | Sum Root to Leaf Numbers

时间：2022-04-02 15:57:28

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 };

 

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