就是找到二叉樹的所有葉子節點，很easy 的DFS 自己的寫法

# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    def leafSimilar(self, root1, root2):
        """
        :type root1: TreeNode
        :type root2: TreeNode
        :rtype: bool
        """
 

        self.array = []
        self.DFS(root1)
        res1 = []+self.array

        self.array=[]
        self.DFS(root2)
        res2 = [] +self.array
        return res1 ==res2

    def DFS(self,root):
        if root.left == None and root.right == None:
            self.array.append(root.val)
        else
 
:
            if root.left!=None:
                self.DFS(root.left)
            if root.right!=None:
                self.DFS(root.right)

大佬的寫法

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

class Solution:
    def leafSimilar(
 
self, root1, root2):
        """
        :type root1: TreeNode
        :type root2: TreeNode
        :rtype: bool
        """
        return self.DFS(root1) == self.DFS(root2)
    def DFS(self,root):
        if root ==None: return []
        #葉子節點
        if root.left ==None and root.right==None:
            return [root.val]
        return self.DFS(root.left) + self.DFS(root.right)

遍歷的時候，用來的儲存遍歷結果的陣列可以不用放在最外層（我寫的那樣），可以讓遍歷函式自己返回（第二種方法），這樣使得程式更具有魯棒性。比如中序遍歷可以寫成如下簡單的形式

    def DFS(self,root):
        if root ==None:return []
        return self.DFS(root.left) +[root.val] + self.DFS(root.right)