LeetCode 129. Sum Root to Leaf Numbers [Medium] [C++] 解題筆記

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.

想法:

    這題熟悉 DFS recursive 的寫法的話還挺簡單的，基本上就是 dfs 做 tree 的 traverse 然後照題目的定義

把每條 root to leaf 的 path 上的值累加起來即可。

完整程式碼:

class Solution {
public:
    int sum;
    void dfs(TreeNode* root, int cur_sum) {
        if (root == nullptr) { return; }
        if (root->left == nullptr && root->right == nullptr) {
            sum += cur_sum*10 + root->val;
            return;
        }
        cur_sum = cur_sum*10 + root->val;
        dfs(root->left, cur_sum);
        dfs(root->right, cur_sum);
        return;
    }
    int sumNumbers(TreeNode* root) {
        sum = 0;
        dfs(root, 0);
        return sum;
    }
};