leetcode 257. Binary Tree Paths

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

For example, given the following binary tree:

 

   1
 /   \
2     3
 \
  5

All root-to-leaf paths are:

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

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# Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None   class Solution(object):     def binaryTreePaths(self, root):         """         :type root: TreeNode         :rtype: List[str]         """         def dfs(node, path, ans):             if not node: return             if not node.left and not node.right: # leaf node                 ans.append(path+str(node.val))                 return                        dfs(node.left, path + str(node.val)+"->", ans)             dfs(node.right, path + str(node.val)+"->", ans)           ans = []         dfs(root, "", ans)         return ans

stack 迭代解法：

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# Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None   class Solution(object):     def binaryTreePaths(self, root):         """         :type root: TreeNode         :rtype: List[str]         """         if not root: return []         stack = [root]         paths = [""]         ans = []         while stack:             node = stack.pop()             path = paths.pop()             if not node.left and not node.right:                 ans.append(path+str(node.val))             if node.right:                 stack.append(node.right)                 paths.append(path+str(node.val)+"->")             if node.left:                 stack.append(node.left)                 paths.append(path+str(node.val)+"->")         return ans

 BFS:（和dfs stack解法就差if顺序）

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# Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None   class Solution(object):     def binaryTreePaths(self, root):         """         :type root: TreeNode         :rtype: List[str]         """         if not root: return []         q = collections.deque([root])         paths = collections.deque([""])         ans = []         while q:             node = q.popleft()             path = paths.popleft()             if not node.left and not node.right:                 ans.append(path+str(node.val))             if node.left:                 q.append(node.left)                 paths.append(path+str(node.val)+"->")             if node.right:                 q.append(node.right)                 paths.append(path+str(node.val)+"->")         return ans

其他解法还有不用helper函数的：

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# Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None   class Solution(object):     def binaryTreePaths(self, root):         """         :type root: TreeNode         :rtype: List[str]         """         if not root: return []         if not root.left and not root.right:             return [str(root.val)]                paths = []         for path in self.binaryTreePaths(root.left):             paths.append(str(root.val) + '->' + path)         for path in self.binaryTreePaths(root.right):             paths.append(str(root.val) + '->' + path)         return paths

 

其他人代码：

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# dfs + stack 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       # bfs + queue def binaryTreePaths2(self, root):     if not root:         return []     res, queue = [], collections.deque([(root, "")])     while queue:         node, ls = queue.popleft()         if not node.left and not node.right:             res.append(ls+str(node.val))         if node.left:             queue.append((node.left, ls+str(node.val)+"->"))         if node.right:             queue.append((node.right, ls+str(node.val)+"->"))     return res       # dfs recursively def binaryTreePaths(self, root):     if not root:         return []     res = []     self.dfs(root, "", res)     return res   def dfs(self, root, ls, res):     if not root.left and not root.right:         res.append(ls+str(root.val))     if root.left:         self.dfs(root.left, ls+str(root.val)+"->", res)     if root.right:         self.dfs(root.right, ls+str(root.val)+"->", res)