129. Sum Root to Leaf Numbers

1.問題

• 給予一個binary tree, 計算所有路徑的總和

2.想法

• 提問

• function header, parameter

• test input

• 觀察

  • DFS, 當到達盡頭時計算總和

  • 總和要用long

• 說明想法

• 測試計算複雜度

3.程式碼

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumNumbers(TreeNode* root) {
        if (!root) {
            return 0;
        }
        
        vector<vector<int>> res; 
        vector<int> record;
        long minVal = 0;
        getPaths(root, record, minVal);
        
        
        return minVal;
    }
private:
    void getPaths(TreeNode* root, vector<int> record, long& sum) {
        if (!root) {
            return;
        } 
        if (!root->left && !root->right) {
            record.push_back(root->val);
            long tmpSum = 0;
            for (int i = 0; i < record.size(); i++) {
               tmpSum += (record[i] * pow (10, record.size() - 1 - i));
            }
            sum += tmpSum;
            return;
        } 
        
        if (root->left) {
            record.push_back(root->val);
            getPaths(root->left, record, sum);
            record.pop_back();
        }
        
        if (root->right) {
            record.push_back(root->val);
            getPaths(root->right, record, sum);
            record.pop_back();
        }
    }
};