257. Binary Tree Paths.md

257. Binary Tree Paths

---

leetcode-cn Daily Challenge on September 4th, 2020.

---

Difficulty : Easy

Related Topics : Tree、DFS

---

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

• mine
  • Java
    • Runtime: 9 ms, faster than 47.07%, Memory Usage: 39.6 MB, less than 81.56% of Java online submissions

      public List<String> binaryTreePaths(TreeNode root) {
          List<String> res = new LinkedList<>();
          if(root == null) return res;
          dfs(res, root, String.valueOf(root.val));
          return res;
      }
      
      void dfs(List<String> res, TreeNode node, String t){
          if(node.left == null && node.right == null){
              res.add(t);
              return;
          }
          if(node.left != null){
              dfs(res, node.left, t + "->" + node.left.val);
          }
          if(node.right != null){
              dfs(res, node.right, t + "->" + node.right.val);
          }
      }
      

---