[LeetCode] 257. Binary Tree Paths_ Easy tag: 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

思路为DFS, 只是append进入stack的时候一并append进入当前的path即可.

1. Constraints
1) can be empty

2. Ideas
DFS     T: O(n)  S: O(n)
1) edge case
2) stack = [(root, str(root.val))]
3) DFS, if leaf, ans.append(path)
4)return ans

3. Codes
3.1) iterable

 class Solution:
     def path(self, root):
         if not root: return []
         ans, stack = [], [(root, str(root.val))]
         while stack
             node, path = stack.pop()
             if not node.left and not node.right:
                 ans.append(path)
             if node.right:
                 stack.append((node.right, path + "->" + node.right.val))
             if node.left:
                 stack.append((node.left, path + "->" + node.left.val))
         return ans

3.2) Recursive

 class Solution:
     def path(self, root):
         def helper(root, path, ans):
             path += str(root.val)
             if not root.left and not root.right:
                 ans.append(path)
             if root.left:
                 stack.append((root.left, path + "->", ans))
             if root.right:
                 stack.append((root.right, path + "->", ans))
         if not root: return []
         ans = []
         helper(root, "", ans)
         return ans

4.. Test cases

1) empty

2)  1

3)

   1
 /   \
2     3
 \
  5