#week18
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"]
分析:
这是一题比较简单的递归
通过递归来求解二叉树路径
题解:
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(vector<string>& result, TreeNode* root, string t) { 13 if (!root->left && !root->right) { 14 result.push_back(t); 15 return ; 16 } 17 18 if (root->left) binaryTreePaths(result, root->left, t + "->" + to_string(root->left->val)); 19 if (root->right) binaryTreePaths(result, root->right, t + "->" + to_string(root->right->val)); 20 21 } 22 vector<string> binaryTreePaths(TreeNode* root) { 23 vector<string> result; 24 if (!root) return result; 25 26 binaryTreePaths(result, root, to_string(root->val)); 27 return result; 28 } 29 };