二叉树的递归解法

标签：nod   tco   http   翻转   for   区别   val   ini   root   

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)

二叉树的递归解法

标签：nod   tco   http   翻转   for   区别   val   ini   root   

原文地址：https://www.cnblogs.com/panzhixiang/p/14928756.html