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Daily Coding Challenge #45

wingkwong profile image wkw ・2 min read

About

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.


/*
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.

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.
*/

/**
 * 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 solve(TreeNode *root, int sum){
        // recursive approach - add *10 for each layer and plus its value
        if(!root) return 0;
        if(!root->left&&!root->right) return sum*10+root->val;
        return solve(root->left, sum*10+root->val)+solve(root->right, sum*10+root->val);
    }
    int sumNumbers(TreeNode* root) {
        return solve(root,0);
    }
};

The source code is available in corresponding repo below. Star and watch for timely updates!

GitHub logo wingkwong / leetcode

πŸ† A Collection of my LeetCode Solutions with Explanations πŸ†

GitHub logo wingkwong / hackerrank

πŸ† A Collection of my HackerRank Solutions with Explanations πŸ†

GitHub logo wingkwong / codeforces

πŸ† A Collection of my Codeforces Solutions with Explanations πŸ†

GitHub logo wingkwong / atcoder

πŸ† A Collection of my AtCoder Solutions with Explanations πŸ†

Discussion

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stjrna profile image
stjrna

hello ;)
if it were an array you can do it like this .. oh yes i didnt compile so figure to fix code .. remember nothing complete.. happy coding

int A[] = {4,9,0,5,1};
int len = sizeof(A)/sizeof(A*); // length
int i=0; int L=0;

while(i<len-3)
{
L = ((i+1)*2)-1;

if (L<len-3) A[L++]+=A[i]*10;
if (L<len-3) A[L]+=A[i++]*10;
}
L=0;
while(len-i) L+=A[i]+A[i++ +1];