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257. Binary Tree Paths

class style def 數組 return ring col ret node

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

For example, given the following binary tree:

   1
 /   2     3
   5

All root-to-leaf paths are:

["1->2->5", "1->3"]
 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6
* TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 void binaryTreePaths(TreeNode*root,vector<string>&res,string t) 13 { 14 if(!root->left&&!root->right) 15 res.push_back(t);
16 if(root->left) 17 binaryTreePaths(root->left,res,t+"->"+to_string(root->left->val)); 18 if(root->right) 19 binaryTreePaths(root->right,res,t+"->"+to_string(root->right->val)); 20 } 21 vector<string> binaryTreePaths(TreeNode* root) {
22 vector<string>res; 23 if(!root) 24 return res; 25 binaryTreePaths(root,res,to_string(root->val)); 26 return res; 27 } 28 };

//如果有子節點,字符串隨著遞歸一直累加,直到沒有子節點,把累加的字符串存進字符串數組

257. Binary Tree Paths