算法描述
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.
示例
Example1:
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.
Example2
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.
个人代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode* root) {
int sum = 0;
int chain = 0;
fun(root,chain,sum);
return sum;
}
int fun(TreeNode* root,int chain,int &sum){
if (root==nullptr) return 1;
chain=chain*10+root->val;
if(fun(root->left,chain,sum)==1 &fun(root->right,chain,sum)==1){
sum=sum+chain;
}
return 2;
}
};
结果
- Accept
- 110/110 cases passed (0 ms)
- Your runtime beats 100 % of cpp submissions
- Your memory usage beats 54.85 % of cpp submissions (12.7 MB)