LeetCode 1022. Sum of Root To Leaf Binary Numbers

🔗 https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers

题目

• 给一个只有 0 和 1 的二叉树
• 从 root 到 leaf 的 path 组成一个二进制数
• 返回这棵树的所有路径和

思路

• dfs 遍历，记录 path 的二进制数
• 匿名函数减少参数的传递，代码很好写！

代码

/**
 * 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 sumRootToLeaf(TreeNode* root) {
        int sum = 0;

        auto dfs = [&](this auto&& self, TreeNode* node, int presum) -> void{
            if (node->right == nullptr && node->left == nullptr) {
                sum = sum + presum * 2 + node->val;
                return;
            }
            if (node->right) self(node->right, presum * 2 + node->val);
            if (node->left) self(node->left, presum * 2 + node->val);
        };
        dfs(root, 0);
        return sum;
        
    }
};