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binary-tree-paths

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Binary Tree Paths – LeetCode 257

Problem

Description

Given a binary tree, return all root-to-leaf paths.

Answer

Original

Code

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/**
 * 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:
    vector<string> Path(TreeNode* root, string s){
        s += "->" + to_string(root->val);
        if(root->left == root->right) return vector<string>({s});
        vector<string> v_s, tmp;
        if(root->left) {
            tmp = Path(root->left,s);
            v_s = std::move(tmp);
        }
        if(root->right) {
            tmp = Path(root->right,s);
            v_s.insert(v_s.end(),tmp.begin(),tmp.end());
        }
        return v_s;
    }
    
    vector<string> binaryTreePaths(TreeNode* root) {
        if(!root) return vector<string>();
        if(root->left != root->right){
            vector<string> v_s, tmp;
            if(root->left) {
                tmp = Path(root->left,to_string(root->val));
                v_s = std::move(tmp);
            }
            if(root->right) {
                tmp = Path(root->right,to_string(root->val));
                v_s.insert(v_s.end(),tmp.begin(),tmp.end());
            }
            return v_s;
        } else
            return vector<string>({to_string(root->val)});
    }
};

思路

使用递归依次遍历然后添加元素，时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$6$ ms，排名$85.23\%$

Better

思路

写的更漂亮，但是并非纯函数，存在性能优化，时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$6$ ms，排名$85.23\%$

Code

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/**
 * 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:
    void binaryTreePaths(vector<string>& result, TreeNode* root, string t) {
        if(root->left == root->right) {
            result.push_back(t);
            return;
        }
        if(root->left) binaryTreePaths(result, root->left, t + "->" + to_string(root->left->val));
        if(root->right) binaryTreePaths(result, root->right, t + "->" + to_string(root->right->val));
    }

    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> result;
        if(!root) return result;
        binaryTreePaths(result, root, to_string(root->val));
        return result;
    }
};