257. Binary Tree Paths

Given the root of a binary tree, return all root-to-leaf paths in any order.

A leaf is a node with no children.

Example 1:

Input: root = [1,2,3,null,5]
Output: ["1->2->5","1->3"]

Example 2:

Input: root = [1]
Output: ["1"]

Constraints:

• The number of nodes in the tree is in the range [1, 100].
• -100 <= Node.val <= 100

recursion

/**
 * 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:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        dfs(root, "", res);
        return res;    
    }

    void dfs(TreeNode* root, string out, vector<string>& res) {
        if(!root->left && !root->right) {
            res.push_back(out + to_string(root->val));
        }
        if(root->left) {
            dfs(root->left, out + to_string(root->val) + "->", res);
        }
        if(root->right) {
            dfs(root->right, out + to_string(root->val) + "->", res);
        }        
    }
};

• 時間複雜度：
  $O(N)$
• 空間複雜度：
  $O(N)$

iteration

/**
 * 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:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        // base case
        if(!root) return res;
        
        // 如果沒有左右子節點，則返回 root
        if(!root->left && !root->right) {
            return {to_string(root->val)};
        }
        
        // 找左側
        for(auto s : binaryTreePaths(root->left)) {
            res.push_back(to_string(root->val) + "->" + s);
        }
        
        // 找右側
        for(auto s : binaryTreePaths(root->right)) {
            res.push_back(to_string(root->val) + "->" + s);
        }

        return res;        
    }
};

• 時間複雜度：
  $O(N)$
• 空間複雜度：
  $O(N)$