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標(biāo)簽（空格分隔）： LeetCode OJ BinaryTree

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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”]

No doubt that we should use recursive method。

/*** 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> binaryTreePaths(TreeNode* root) {if (root == NULL){vector<string> empty;return empty;} if (root->left == NULL && root->right == NULL){ // leafvector<string> path;path.push_back(std::to_string(root->val));return path;}vector<string> allPaths = combine(binaryTreePaths(root->left), binaryTreePaths(root->right)); // combine all subpathsvector<string> result;for (vector<string>::iterator it = allPaths.begin(); it != allPaths.end(); ++it){result.push_back(std::to_string(root->val) + "->" + *it); // add current node to the paths}return result;}vector<string> combine(vector<string> left, vector<string> right){ // this is how to combine two vectorsvector<string> result;result.reserve(left.size() + right.size() ); // preallocate memoryresult.insert( result.end(), left.begin(), left.end() );result.insert( result.end(), right.begin(), right.end() );return result;} }; 《新程序員》：云原生和全面數(shù)字化實(shí)踐50位技術(shù)專家共同創(chuàng)作，文字、視頻、音頻交互閱讀

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