【代码随想录day15】翻转二叉树-CSDN博客

本文链接：https://blog.csdn.net/doubleguy/article/details/131828725

题目

给你一棵二叉树的根节点 root ，翻转这棵二叉树，并返回其根节点。

示例 1：

输入：root = [4,2,7,1,3,6,9]
输出：[4,7,2,9,6,3,1]
示例 2：

输入：root = [2,1,3]
输出：[2,3,1]
示例 3：

输入：root = []
输出：[]

提示：

树中节点数目范围在 [0, 100] 内
-100 <= Node.val <= 100

思路

整体思想是利用遍历过程，在遍历时将每个节点的左右元素进行交换。所以可以用三种递归遍历和非递归方式进行遍历。但是这里要注意中序递归遍历会有点问题，尽量用前序和后序遍历吧。

递归前序遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        def solve(root):
            if not root:
                return 
            root.left, root.right = root.right, root.left
            solve(root.left)
            solve(root.right)
        solve(root)
        return root

递归后序遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        def solve(root):
            if not root:
                return
            solve(root.left)
            solve(root.right)
            root.left, root.right = root.right, root.left
        solve(root)
        return root

非递归前序遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return 
        def solve(root):
            stack = [root]
            while stack:
                now = stack.pop()
                now.left, now.right = now.right, now.left
                if now.right:
                    stack.append(now.right)
                if now.left:
                    stack.append(now.left)
        solve(root)
        return root

非递归后续遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return 
        def solve(root):
            stack = [root]
            while stack:
                now = stack.pop()
                if now.left:
                    stack.append(now.left)
                if now.right:
                    stack.append(now.right)
                now.left, now.right = now.right, now.left
        solve(root)
        return root

补充一下层序遍历代码：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return 
        def solve(root):
            q = deque([root])
            while q:
                now = q.popleft()
                if now.left:
                    q.append(now.left)
                if now.right:
                    q.append(now.right)
                now.left, now.right = now.right, now.left
        solve(root)
        return root