二叉树的根到叶子几点之和

输入：root = [1,2,3]
输出：25
解释：
从根到叶子节点路径 1->2 代表数字 12
从根到叶子节点路径 1->3 代表数字 13
因此，数字总和 = 12 + 13 = 25

输入：root = [1,0,1,0,1,0,1]
输出：22
解释：(100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22  

129. 求根节点到叶节点数字之和 https://leetcode.cn/problems/sum-root-to-leaf-numbers/

/**
 * 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) {
        return getNum(root,0);
    }

    int getNum(TreeNode* root,int num) {
        if(root==nullptr) {
            return 0;
        }

        num = num*10+root->val;
        if(root->left==nullptr&&root->right==nullptr) {
            return num;
        }

        return getNum(root->left,num)+getNum(root->right,num);

    }
};

1022. 从根到叶的二进制数之和 https://leetcode.cn/problems/sum-of-root-to-leaf-binary-numbers/

class Solution {
public:
    int sumRootToLeaf(TreeNode* root) {
        return getNum(root,0);
    }

    int getNum(TreeNode* root,int num) {
        if(root==nullptr) {
            return 0;
        }

        num = num*2+root->val;
        if(root->left==nullptr&&root->right==nullptr) {
            return num;
        }

        return getNum(root->left,num)+getNum(root->right,num);

    }
};

两个题目本质上是一道内容：

唯一的修改点是：

 

        num = num*2+root->val;

        num = num*10+root->val;