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The basic idea is to start from root and add it to the current path, then we recursively visit its left and right subtrees if they exist; otherwise, we have reached a leaf node, so just add the current path to the result paths.
The code is as follows.
1 class Solution { 2 public: 3 vector<string> binaryTreePaths(TreeNode* root) { 4 vector<string> paths; 5 string path; 6 treePaths(root, path, paths); 7 return paths; 8 } 9 private: 10 void treePaths(TreeNode* node, string path, vector<string>& paths) { 11 if (!node) return; 12 path += path.empty() ? to_string(node -> val) : "->" + to_string(node -> val); 13 if (!(node -> left) && !(node -> right)) { 14 paths.push_back(path); 15 return; 16 } 17 if (node -> left) treePaths(node -> left, path, paths); 18 if (node -> right) treePaths(node -> right, path, paths); 19 } 20 };
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4733742.html
