代码随想录 - Day21 - 翻转二叉树

代码随想录 - Day21 - 翻转二叉树

226. 翻转二叉树

翻转二叉树的题目用前序遍历、后序遍历、层序遍历都可以做，但递归的中序遍历不能用（使用递归的中序遍历，某些节点的左右孩子会翻转两次）。

• 递归法（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 None
        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root

迭代法：前序遍历

class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
        stack = [root]
        while stack:
            node = stack.pop()
            node.left, node.right = node.right, node.left
            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)
        return root

递归法：中序遍历
两次self.invertTree(root.left)避免节点左右孩子翻转两次

class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
        self.invertTree(root.left)
        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        return root

迭代法：中序遍历
两次if node.left: stack.append(node.left)避免节点左右孩子翻转两次

class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
        stack = [root]
        while stack:
            node = stack.pop()
            if node.left:
                stack.append(node.left)
            node.left, node.right = node.right, node.left
            if node.left:
                stack.append(node.left)
        return root

递归法：后序遍历

class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
        
        self.invertTree(root.left)
        self.invertTree(root.right)
        root.left, root.right = root.right, root.left
        return root

迭代法：后序遍历

class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
        stack = [root]
        while stack:
            node = stack.pop()
            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)
            node.left, node.right = node.right, node.left
        return root

总结

针对二叉树的问题，解题之前一定要想清楚究竟是前中后序遍历，还是层序遍历。

二叉树解题的大忌就是自己稀里糊涂的过了，但不知道自己是怎么遍历的。