文章目录
一、题目
Given the root of a binary tree, return all root-to-leaf paths in any order.
A leaf is a node with no children.
Example 1:
Input: root = [1,2,3,null,5]
Output: ["1->2->5","1->3"]
Example 2:
Input: root = [1]
Output: ["1"]
Constraints:
The number of nodes in the tree is in the range [1, 100].
-100 <= Node.val <= 100
二、题解
前序遍历+回溯
cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void getPath(TreeNode* root,vector<int>& path,vector<string>& res){
path.push_back(root->val);
if(root->left == nullptr && root->right == nullptr){
string s;
for(int i = 0;i < path.size();i++){
s += to_string(path[i]);
if(i != path.size() - 1) s += "->";
}
res.push_back(s);
}
if(root->left){
getPath(root->left,path,res);
path.pop_back();
}
if(root->right){
getPath(root->right,path,res);
path.pop_back();
}
}
vector<string> binaryTreePaths(TreeNode* root) {
vector<int> path;
vector<string> res;
getPath(root,path,res);
return res;
}
};