def buildTree(self, inorder: [int], postorder: [int]) -> TreeNode:
"""
Solution:
1. 如果 inorder 列表为空，返回 None
2. 如果 inorder 长度为 1，直接返回
3. 由 postorder[-1] 拿到根节点
4. 找到根节点在 inorder[] 中的位置
5. 根据 inorder 和 postorder 列表等长的特性
1. 递归左子树
2. 赌鬼右子树
:param inorder:
:param postorder:
:return:
"""
if not inorder:
return None
if len(inorder) == 1:
return TreeNode(inorder[0])
root = TreeNode(postorder[-1])
sep = inorder.index(postorder[-1])
if sep == 0:
root.right = self.buildTree(inorder[1:], postorder[:-1])
elif sep == len(inorder) - 1:
root.left = self.buildTree(inorder[:-1], postorder[:-1])
else:
root.left = self.buildTree(inorder[:sep], postorder[:sep])
root.right = self.buildTree(inorder[(sep + 1):], postorder[sep:-1])
return root
def invertTree(self, root: TreeNode) -> TreeNode:
if root:
# 保存右节点的信息
right = root.right
# 右节点递归翻转
root.right = self.invertTree(root.left)
# 左节点递归翻转
root.left = self.invertTree(right)
return root
else:
return None
def mergeTree2(t1: TreeNode, t2: TreeNode):
if not t1 and not t2:
return
if not t1:
t1 = t2
return
if not t2:
return
t1.val += t2.val
if t1.left and t2.left:
mergeTree2(t1.left, t2.left)
elif t2.left:
t1.left = t2.left
if t1.right and t2.right:
mergeTree2(t1.right, t2.right)
elif t2.right:
t1.right = t2.right