[每日 LeetCode] 1022. Sum of Root To Leaf Binary Numbers

原文链接 [每日 LeetCode] 1022. Sum of Root To Leaf Binary Numbers

Description:

Given a binary tree, each node has value 0 or 1.  Each root-to-leaf path represents a binary number starting with the most significant bit.  For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is 13.

For all leaves in the tree, consider the numbers represented by the path from the root to that leaf.

Return the sum of these numbers.

Example 1:

Input: [1,0,1,0,1,0,1]
Output: 22
Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22

Note:

1. The number of nodes in the tree is between 1 and 1000.
2. node.val is 0 or 1.
3. The answer will not exceed 2^31 - 1.

思路：本题要求二叉树中所有路径的二进制编码之和，涉及到二叉树的遍历。在此使用 DFS，遍历每层时记录下当前的数值之和，直到叶子节点。注意初始化二叉树的方式，默认初始化为完全二叉树。

C++ 代码

/**
 * 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 sumRootToLeaf(TreeNode* root) {
        if (!root)
            return 0;
        int sum = 0;
        DFS(root, 0, sum);
        return sum;
    }
    void DFS(TreeNode *root, int cur_sum, int& sum){
        if (!root)
            return;
        cur_sum = 2 * cur_sum + root->val;
        if (!root->left && !root->right)
            sum += cur_sum;
        if (root->left)
            DFS(root->left, cur_sum, sum);
        if (root->right)
            DFS(root->right, cur_sum, sum);
    }
};

运行时间：8ms

运行内存：17.2M