129. Sum Root to Leaf Numbers

preorder O(n) time O(h) space

/**
 * 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, int num = 0) {
        return sum(root, 0);
    }

private:
    long sum(TreeNode *root, long val) { // 返回这棵树所有Root to Leaf Numbers的和
        if (!root) return 0;
        val = val * 10 + root->val;
        if (!root->left && !root->right) return val; // 叶节点要单独处理
        return sum(root->left, val) + sum(root->right, val);
    }
};