257. 二叉樹的所有路徑

給定一個二叉樹，返回所有從根節點到葉子節點的路徑。

說明: 葉子節點是指沒有子節點的節點。

示例:
輸入:
   1
 /   \
2     3
 \
  5
輸出: ["1->2->5", "1->3"]

解釋: 所有根節點到葉子節點的路徑爲: 1->2->5, 1->3
算法一：BFS

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    def binaryTreePaths(self, root: TreeNode):
        from collections import deque
        if not root: return []
        res = []
        queue = deque()
        queue.appendleft([root, []])
        while queue:
            node, tmp = queue.pop()###最重要的就是這個temp，噹噹前節點左存在的時候，把當前節點的值記錄下來。這樣一直記錄就能夠記錄到葉子節點。葉子節點左右爲空。返回res
            if not node.left and not node.right:
                res.append("->".join(tmp + [str(node.val)]))
            if node.left:
                queue.appendleft([node.left, tmp + [str(node.val)]])
            if node.right:
                queue.appendleft([node.right, tmp + [str(node.val)]])
        return res
root = TreeNode(1)
a = TreeNode(2)
b = TreeNode(3)
c = TreeNode(5)
root.left = a
root.right = b
a.right = c
print(Solution().binaryTreePaths(root))

算法二：DFS

class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        if not root: return []
        res = []
        def helper(root, tmp):
            if not root.left and not root.right:res.append(tmp + [str(root.val)]) 
            if root.left:helper(root.left, tmp + [str(root.val)])
            if root.right:helper(root.right, tmp + [str(root.val)])
        helper(root, [])
        return ["->".join(a) for a in res]