给定一个二叉树,返回所有从根节点到叶子节点的路径。
说明: 叶子节点是指没有子节点的节点。
示例:
输入:
1 / \ 2 3 \ 5
输出: ["1->2->5", "1->3"]
解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3
解法一:
/** * 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; dfs(root, res, ""); return res; } void dfs(TreeNode *root, vector<string> &res, string curr) { if(!root) return; curr += to_string(root->val); if(root->left == NULL && root->right == NULL) { res.push_back(curr); return; } dfs(root->left, res, curr + "->"); dfs(root->right, res, curr + "->"); } };