257 Binary Tree Paths¶

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

Solution: Recursion¶

Time complexity: O(n) Space complexity: O(n) / output can be O(n^2)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new LinkedList<String>();
        process(root, "", result);
        return result;
    }
    private void process(TreeNode node, String curr, List<String> result) {
        if (node == null) {
            return;
        }
        String next = curr + "->" + node.val;
        if (node.left == null && node.right == null) {
            result.add(next.substring(2));
        }
        process(node.left, next, result);
        process(node.right, next, result);
    }
}

class Solution {
public:
  vector<string> binaryTreePaths(TreeNode* root) {
    vector<string> ans;
    string s;
    function<void(TreeNode*, int)> preorder = [&](TreeNode* node, int l) {
      if (!node) return;
      s += (l > 0 ? "->" : "") + to_string(node->val);
      if (!node->left && !node->right)
        ans.push_back(s);
      preorder(node->left, s.size());
      preorder(node->right, s.size());
      while (s.size() != l) s.pop_back();
    };
    preorder(root, 0);
    return ans;
  }
};