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# Binary Tree Paths LeetCode Solution

## Problem – Binary Tree Paths LeetCode Solution

Given the `root` of a binary tree, return all root-to-leaf paths in any order.

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 = 
Output: ["1"]
``````

Constraints:

• The number of nodes in the tree is in the range `[1, 100]`.
• `-100 <= Node.val <= 100`

## Binary Tree Paths LeetCode Solution in Java

``````public List<String> binaryTreePaths(TreeNode root) {
List<String> answer = new ArrayList<String>();
if (root != null) searchBT(root, "", answer);
}
private void searchBT(TreeNode root, String path, List<String> answer) {
if (root.left == null && root.right == null) answer.add(path + root.val);
if (root.left != null) searchBT(root.left, path + root.val + "->", answer);
if (root.right != null) searchBT(root.right, path + root.val + "->", answer);
}
``````

## Binary Tree Paths LeetCode Solution in Python

``````    def binaryTreePaths1(self, root):
if not root:
return []
res, stack = [], [(root, "")]
while stack:
node, ls = stack.pop()
if not node.left and not node.right:
res.append(ls+str(node.val))
if node.right:
stack.append((node.right, ls+str(node.val)+"->"))
if node.left:
stack.append((node.left, ls+str(node.val)+"->"))
return res
``````

## Binary Tree Paths LeetCode Solution in C++

``````void binaryTreePaths(vector<string>& result, TreeNode* root, string t) {
if(!root->left && !root->right) {
result.push_back(t);
return;
}

if(root->left) binaryTreePaths(result, root->left, t + "->" + to_string(root->left->val));
if(root->right) binaryTreePaths(result, root->right, t + "->" + to_string(root->right->val));
}

vector<string> binaryTreePaths(TreeNode* root) {
vector<string> result;
if(!root) return result;

binaryTreePaths(result, root, to_string(root->val));
return result;
}
``````
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