Description
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
Solutions
Solution 1
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23 | # Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
def dfs(root: Optional[TreeNode]):
if root is None:
return
t.append(str(root.val))
if root.left is None and root.right is None:
ans.append("->".join(t))
else:
dfs(root.left)
dfs(root.right)
t.pop()
ans = []
t = []
dfs(root)
return ans
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38 | /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private List<String> ans = new ArrayList<>();
private List<String> t = new ArrayList<>();
public List<String> binaryTreePaths(TreeNode root) {
dfs(root);
return ans;
}
private void dfs(TreeNode root) {
if (root == null) {
return;
}
t.add(root.val + "");
if (root.left == null && root.right == null) {
ans.add(String.join("->", t));
} else {
dfs(root.left);
dfs(root.right);
}
t.remove(t.size() - 1);
}
}
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44 | /**
* 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:
vector<string> binaryTreePaths(TreeNode* root) {
vector<string> ans;
vector<string> t;
function<void(TreeNode*)> dfs = [&](TreeNode* root) {
if (!root) {
return;
}
t.push_back(to_string(root->val));
if (!root->left && !root->right) {
ans.push_back(join(t));
} else {
dfs(root->left);
dfs(root->right);
}
t.pop_back();
};
dfs(root);
return ans;
}
string join(vector<string>& t, string sep = "->") {
string ans;
for (int i = 0; i < t.size(); ++i) {
if (i > 0) {
ans += sep;
}
ans += t[i];
}
return ans;
}
};
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27 | /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func binaryTreePaths(<
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