257. Binary Tree Paths
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题目: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"]
题意:
给定一棵二叉树,返回所有从根节点到叶子节点的路径。
转载地址:http://www.cnblogs.com/grandyang/p/4738031.html
思路一:
采用递归实现(DFS),只需要每次迭代每个节点,判断是否有左右子树,如果有,则添加->符号之后,将子节点放到新路径之后,继续对左右子树进行递归遍历。
代码:C++版:4ms
/** * 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) { vector<string> res; if (root) dfs(root, "", res); return res; } void dfs(TreeNode *root, string out, vector<string> &res) { out += to_string(root->val); if (!root->left && !root->right) { res.push_back(out); } else { if (root->left) dfs(root->left, out + "->", res); if (root->right) dfs(root->right, out + "->", res); } }};
思路二:
另一种递归方法,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dfs函数,核心思想和上面没有区别。
代码:C++版:4ms
/** * 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) 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; }};
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- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
- 257. Binary Tree Paths
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