[LeetCode] 257. Binary Tree Paths 二叉树路径
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"]
这道题给我们一个二叉树,让返回所有根到叶节点的路径,跟之前那道 Path Sum II 很类似,比那道稍微简单一些,不需要计算路径和,只需要无脑返回所有的路径即可,那么思路还是用递归来解,博主之前就强调过,玩树的题目,十有八九都是递归,而递归的核心就是不停的 DFS 到叶结点,然后在回溯回去。在递归函数中,当遇到叶结点的时候,即没有左右子结点,那么此时一条完整的路径已经形成了,加上当前的叶结点后存入结果 res 中,然后回溯。注意这里结果 res 需要 reference,而 out 是不需要引用的,不然回溯回去还要删除新添加的结点,很麻烦。为了减少判断空结点的步骤,我们在调用递归函数之前都检验一下非空即可,代码而很简洁,参见如下:
解法一:
class Solution { public: vector<string> binaryTreePaths(TreeNode* root) { vector<string> res; if (root) helper(root, "", res); return res; } void helper(TreeNode* node, string out, vector<string>& res) { if (!node->left && !node->right) res.push_back(out + to_string(node->val)); if (node->left) helper(node->left, out + to_string(node->val) + "->", res); if (node->right) helper(node->right, out + to_string(node->val) + "->", res); } };
下面再来看一种递归的方法,这个方法直接在一个函数中完成递归调用,不需要另写一个 helper 函数,核心思想和上面没有区别,参见代码如下:
解法二:
class Solution { public: vector<string> binaryTreePaths(TreeNode* root) { if (!root) return {}; if (!root->left && !root->right) return {to_string(root->val)}; vector<string> left = binaryTreePaths(root->left); vector<string> right = binaryTreePaths(root->right); left.insert(left.end(), right.begin(), right.end()); for (auto &a : left) { a = to_string(root->val) + "->" + a; } return left; } };
还是递归写法,从论坛中扒下来的解法,核心思路都一样啦,写法各有不同而已,参见代码如下:
解法三:
class Solution { public: vector<string> binaryTreePaths(TreeNode* root) { if (!root) return {}; if (!root->left && !root->right) return {to_string(root->val)}; vector<string> res; for (string str : binaryTreePaths(root->left)) { res.push_back(to_string(root->val) + "->" + str); } for (string str : binaryTreePaths(root->right)) { res.push_back(to_string(root->val) + "->" + str); } return res; } };
Github 同步地址:
https://github.com/grandyang/leetcode/issues/257
类似题目:
参考资料:
https://leetcode.com/problems/binary-tree-paths/