二叉树的递归解法
Posted 叶夏
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贴上大佬的博客地址：https://labuladong.gitee.io/algo/2/18/20/
这一部分总得来说比较简单，注意边界值的判断就行了。下面是贴上实际的代码
"""
翻转二叉树
https://leetcode-cn.com/problems/invert-binary-tree/
"""
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: TreeNode) -> TreeNode:
"""
注意判断边界值，
在子节点递归调用翻转函数
"""
if root == None:
return root
tmp = root.left
root.left = root.right
root.right = tmp
self.invertTree(root.left)
self.invertTree(root.right)
return root
"""
填充每个节点的下一个右侧节点
https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/
"""
# Definition for a Node.
class Node:
def __init__(self, val: int = 0, left: \'Node\' = None, right: \'Node\' = None, next: \'Node\' = None):
self.val = val
self.left = left
self.right = right
self.next = next
class Solution:
"""
注意对比与226 翻转二叉树的区别，这里的递归函数中传入了两个参数（两个节点）
"""
def connect(self, root: \'Node\') -> \'Node\':
if root == None:
return root
self.connectTwoNode(root.left, root.right)
return root
def connectTwoNode(self, left, right):
# 编辑判断要放在这里做，在connect中做没有用，因为递归调用的是这个函数
if left == None or right == None:
return None
left.next = right
self.connectTwoNode(left.left, left.right)
self.connectTwoNode(left.right, right.left)
self.connectTwoNode(right.left, right.right)
"""
填充每个节点的下一个右侧节点
https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/
"""
# Definition for a Node.
class Node:
def __init__(self, val: int = 0, left: \'Node\' = None, right: \'Node\' = None, next: \'Node\' = None):
self.val = val
self.left = left
self.right = right
self.next = next
class Solution:
"""
注意对比与226 翻转二叉树的区别，这里的递归函数中传入了两个参数（两个节点）
"""
def connect(self, root: \'Node\') -> \'Node\':
if root == None:
return root
self.connectTwoNode(root.left, root.right)
return root
def connectTwoNode(self, left, right):
# 编辑判断要放在这里做，在connect中做没有用，因为递归调用的是这个函数
if left == None or right == None:
return None
left.next = right
self.connectTwoNode(left.left, left.right)
self.connectTwoNode(left.right, right.left)
self.connectTwoNode(right.left, right.right)
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