257. 二叉树的所有路径

一、递归

前序遍历，递归终止条件为左右孩子都为空，说明当前结点为叶子结点，然后将路径添加至结果中，单层递归逻辑同前序遍历，在递归后回溯，每个递归都要有对应的回溯

class Solution {
public:
    void dfs(TreeNode* node,vector<int> &path,vector<string> &result)
    {
        //中
        path.push_back(node->val);
        if(node->left==NULL && node->right==NULL)
        {
            string s;
            for(int i =0;i<path.size()-1;i++)
            {
                s += to_string(path[i]);
                s += "->";
            }
            s += to_string(path[path.size()-1]);
            result.push_back(s);
        }
        //左
        if(node->left)
        {
            dfs(node->left,path,result);
            path.pop_back();
        }
        //右
        if(node->right)
        {
            dfs(node->right,path,result);
            path.pop_back();
        }
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        if(root==NULL)return {};
        vector<string> result;
        vector<int> path;
        dfs(root,path,result);
        return result;
    }
};

简化

通过函数的参数进行回溯操作

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;
    }
};