吃肉的刷题记录8-二叉树，链表

二叉树转单链表

二叉树转换为单链表，展开后的单链表应该与二叉树先序遍历顺序相同。

1

/ \

2 5

/ \ \

3 4 6

简单方法一：

class TreeNode:
    def __init__(self,x):
        self.val = x
        self.left = None
        self.right = None
class ListNode:
    def __init__(self,x):
        self.val = x
        self.next = None

def pre(root):
    if root:
        list_tree.append(root.val)
        pre(root.left)
        pre(root.right)
def creatListNode(l):
    if l==None:
        return None
    head = ListNode(l[0])
    curr = head
    for i in range(1,len(l)):
        curr.next = ListNode(l[i])
        curr = curr.next
    return head


root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(5)
root.left.left = TreeNode(3)
root.left.right = TreeNode(4)
root.right.right = TreeNode(6)

list_tree = []
pre(root)
print(list_tree)
ret_ListNode = creatListNode(list_tree)
while ret_ListNode:
    print(ret_ListNode.val)
    ret_ListNode = ret_ListNode.next

优化方法二、

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

def flatten(root: TreeNode) -> ListNode:
    if not root:
        return None
    
    # 创建单链表的头节点
    head = ListNode(root.val)
    
    # 递归处理左子树和右子树
    left_list = flatten(root.left)
    right_list = flatten(root.right)
    
    # 将左子树链表连接到当前节点后面
    head.next = left_list
    # 找到左子树链表的最后一个节点，并将右子树链表连接到其后
    last = head
    while last.next:
        last = last.next
    last.next = right_list
    
    return head

def flatten_tree_to_list(root: TreeNode) -> ListNode:
    return flatten(root)

# 构建示例二叉树
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(5)
root.left.left = TreeNode(3)
root.left.right = TreeNode(4)
root.right.right = TreeNode(6)

# 转换为单链表并返回头节点
list_head = flatten_tree_to_list(root)

# 输出单链表的值，以验证结果
result = []
while list_head:
    result.append(list_head.val)
    list_head = list_head.next
result

中序后序遍历构造二叉树

描述
给定一个二叉树的中序与后序遍历结果，请你根据两个序列构造符合这两个序列的二叉树。

例如输入[2,1,4,3,5],[2,4,5,3,1]时，
根据中序遍历的结果[2,1,4,3,5]和后序遍历的结果[2,4,5,3,1]可构造出二叉树{1,2,3,#,#,4,5}，如下图所示：

示例1
输入：
[1],[1]
返回值：
{1}

示例2
输入：
[2,1,4,3,5],[2,4,5,3,1]
返回值：
{1,2,3,#,#,4,5}

思路：

中序遍历: 左 -> 根 -> 又 后序遍历: 左 -> 右 -> 根 则后序遍历的最后一个节点为根节点,该节点在中序数组中的索引 idx 的左侧即为左子树[0:idx],右侧为右子树 [idx + 1:]，该索引代表了左子树的长度,在后序数组中左子树为[0:idx],右子树为[idx:-1],从上至下建立树

代码

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param inorder int整型一维数组 中序遍历序列
# @param postorder int整型一维数组 后序遍历序列
# @return TreeNode类
#
class Solution:
    def buildTree(self , inorder: List[int], postorder: List[int]) -> TreeNode:
        # write code here
        if not inorder or not postorder:
            return None
        root = TreeNode(postorder[-1])
        idx = inorder.index(postorder[-1])
        root.left = self.buildTree(inorder[:idx],postorder[:idx])
        root.right = self.buildTree(inorder[idx+1:],postorder[idx:-1])
        return root