binary-tree - “if right and left: return root”和“return right or left”如何帮助我们找到最不共同的祖先？

Question

class TreeNode:
   def __init__(self, data, left = None, right = None):
      self.data = data
      self.left = left
      self.right = right
def insert(temp,data):
   que = []
   que.append(temp)
   while (len(que)):
      temp = que[0]
      que.pop(0)
      if (not temp.left):
         if data is not None:
            temp.left = TreeNode(data)
         else:
            temp.left = TreeNode(0)
         break
      else:
         que.append(temp.left)
         if (not temp.right):
            if data is not None:
               temp.right = TreeNode(data)
            else:
               temp.right = TreeNode(0)
            break
         else:
            que.append(temp.right)
def make_tree(elements):
   Tree = TreeNode(elements[0])
   for element in elements[1:]:
      insert(Tree, element)
   return Tree
class Solution(object):
   def lowestCommonAncestor(self, root, p, q):
      if not root:
         return None
      if root.data == p or root.data ==q:
         return root
      left = self.lowestCommonAncestor(root.left, p, q)
      right = self.lowestCommonAncestor(root.right, p, q)
      **if right and left:
         return root**
      *return right or left*
ob1 = Solution()

---

tree = make_tree([3,5,1,6,2,0,8,None,None,7,4])
print(ob1.lowestCommonAncestor(tree, 5, 1).data)

当我们想要返回单个节点而不是两个节点时，为什么要返回右或左？

我意识到即使左右不是同一个TreeNode，布尔值“左右”也会返回true。此外，没有编写比较函数。那有必要吗？

Answer 1

return right or leftright如果不是，则返回None，否则，它检查left并返回left，如果不是None。如果left两者right都None返回None。因此，只返回一个节点 or None。

if right and left只有True当两者left都不right是None