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 = [1]
Output: ["1"]


  • 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);
    return 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:
            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) {

    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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