🔗 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
cpp
/**
* 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;
}
};