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 + "->");
}
};