力扣257. 二叉树的所有路径

/**
 * 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:
    string makePath(const vector<int>& path) {
        string path_str = to_string(path[0]);
        for(int i = 1; i < path.size(); i++) {
            path_str = path_str + "->" + to_string(path[i]);
        }
        return path_str;
    }

    void getPath(TreeNode* node, vector<int>& path, vector<string>& result) {
        path.push_back(node->val);
        if(node->left == nullptr && node->right == nullptr) {
            result.push_back(makePath(path));
            return;
        }
        if(node->left) {
            getPath(node->left, path, result);
            path.pop_back();
        }
        if(node->right) {
            getPath(node->right, path, result);
            path.pop_back();
        }
    }

    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> result;
        vector<int> path;
        getPath(root, path, result);
        return result;
    }
};

精简版

class Solution {
private:

    void traversal(TreeNode* cur, string path, vector<string>& result) {
        path += to_string(cur->val); // 中
        if (cur->left == NULL && cur->right == NULL) {
            result.push_back(path);
            return;
        }
        if (cur->left) traversal(cur->left, path + "->", result); // 左
        if (cur->right) traversal(cur->right, path + "->", result); // 右
    }

public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> result;
        string path;
        if (root == NULL) return result;
        traversal(root, path, result);
        return result;

    }
};