Leetcode 129: Sum Root to Leaf Numbers (树遍历题)

129. Sum Root to Leaf Numbers
     Medium
     7.3K
     116
     Companies
     You are given the root of a binary tree containing digits from 0 to 9 only.

Each root-to-leaf path in the tree represents a number.

For example, the root-to-leaf path 1 -> 2 -> 3 represents the number 123.
Return the total sum of all root-to-leaf numbers. Test cases are generated so that the answer will fit in a 32-bit integer.

A leaf node is a node with no children.

Example 1:

Input: root = [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: root = [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.

Constraints:

The number of nodes in the tree is in the range [1, 1000].
0 <= Node.val <= 9
The depth of the tree will not exceed 10.

解法1：

/**
 * 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) {
        if (!root) return 0;
        string path = to_string(root->val);
        helper(root, path);
        return sum;
    }
private:
    int sum = 0;
    void helper(TreeNode *root, string path) {
        if (!root) return;
        if (!root->left && !root->right) {
            sum += stoi(path);
            return;
        }
        if (root->left) helper(root->left, path + to_string(root->left->val));
        if (root->right) helper(root->right, path + to_string(root->right->val));
        return;
    }
};

写成这样也可以

/**
 * 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) {
        if (!root) return 0;
        string path = "";
        helper(root, path);
        return sum;
    }
private:
    int sum = 0;
    void helper(TreeNode *root, string path) {
        if (!root) return;
        path += to_string(root->val);
        if (!root->left && !root->right) {
            sum += stoi(path);
            return;
        }

        if (root->left) helper(root->left, path);
        if (root->right) helper(root->right, path);
        return;
    }
};