Leetcode | 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.

For example,

1
/
2 3
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

这道题就很简单了。反正就是递归，每遍历到一个点，就把前面计算到的数*10+当前数，到叶子结点的时候就把数加一下。

叶子结点就是左右结点都为NULL。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumNumbers(TreeNode *root) {
        int sum = 0;   
        recursive(root, 0, sum);
        return sum;
    }
    
    void recursive(TreeNode* root, int value, int& sum) {
        if (root == NULL) {
            return;
        }
        int v = value * 10 + root->val;
        if (root->left == NULL && root->right == NULL) {
            sum += v;
        }
        recursive(root->left, v, sum);
        recursive(root->right, v, sum);
    }
};

 第三遍刷，写了另外一种。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumNumbers(TreeNode *root) {
        return recurse(root, 0);
    }
    
    int recurse(TreeNode *root, int sum) {
        if (root == NULL) return 0;
        if (root->left == NULL && root->right == NULL) return sum * 10 + root->val;
        int left = recurse(root->left, sum * 10 + root->val);
        int right = recurse(root->right, sum * 10 + root->val);
        return left + right;
    }
};