leetcode236 Lowest Common Ancestor of a Binary Tree

 """
 Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
 According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
 Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]
 Example 1:
 Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
 Output: 3
 Explanation: The LCA of nodes 5 and 1 is 3.
 Example 2:
 Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
 Output: 5
 Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
 """
 """
 本题两种解法。第一种非递归的方法
 用queue来存储结点遍历
 用dict{root, parent[root]}来存储每个结点的父亲结点
 然后用一个set存储p的所有父辈结点
 再遍历q的每个父亲结点查找是否再set中
 如果找到即为p,q结点的最近公共祖先
 """
 class TreeNode:
     def __init__(self, x):
         self.val = x
         self.left = None
         self.right = None

 class Solution1:
     def lowestCommonAncestor(self, root, p, q):
         queue = [root]       #用来层次遍历
         parent = {root: None} #用dict存储父亲结点
         while queue:
             node = queue.pop()
             if node.left:
                 parent[node.left] = node  #存父亲结点
                 queue.append(node.left)   #入队
             if node.right:
                 parent[node.right] = node
                 queue.append(node.right)
         res = set() #set集合是一个无序不重复元素的序列
         while p:    #res=() 这是把res定义为tuple,tuple是只能查看的list
             res.add(p)    #将p的所有父辈结点放入set里
             p = parent[p]
         while q not in res: #q向上找到相同的父亲结点
             q = parent[q]
         return q

 """
 第二种是递归写法：没有理解
 传送门：https://blog.csdn.net/qq_17550379/article/details/95903394
 树型问题首先考虑递归，对于每个树中的p和q只会有一下几种情况
 1. p在左子树中，q在右子树中
 2. q在左子树中，p在右子树中
 3. p和q都在左子树中
 4. p和q都在右子树中
 对于第一种和第二种情况很简单，p和q的最近公共祖先就是root。
 对于第三种情况我们只需递归向左子树找，第四种情况我们只需递归向右子树找。接着思考边界情况。
 当p==root or q==root的时候，我们返回root即可（因为要找最近公共祖先，继续遍历的话，就不可能是其祖先了）。
 那么这里就有一个迷惑人的地方了，请认真思考第一种和第二种情况下，左右子树的递归返回结果是什么？就是p和q。
 """

 class Solution2:
     def lowestCommonAncestor(self, root, p, q):
         if not root or root == p or root == q:
             return root
         left = self.lowestCommonAncestor(root.left, p, q)
         right = self.lowestCommonAncestor(root.right, p, q)
         if left and right:
             return root
         return left if left else right