二叉树与链表

二叉树

1. 建立二叉树

class Node:
    def __init__(self, data, left=None, right=None):
        self.data = data
        self.left = left
        self.right = right

tree = Node(1, Node(3, Node(7, Node(0)), Node(6)), Node(2, Node(5)))

2. 广度遍历与深度遍历

广度遍历：

def lookup(tree):
    if not tree:
        return []
    l = []
    stack = [tree, ]
    while stack:
        node = stack.pop(0)
        l.append(node.data)
        if node.left:
            stack.append(node.left)
        if node.right:
            stack.append(node.right)
    return l

深度遍历：

前序遍历：根，左子树，右子树

def pre(tree):
    l = []

    def deep(tree):
        if not tree:
            return
        l.append(tree.data)
        deep(tree.left)
        deep(tree.right)
    deep(tree)
    return l

中序遍历：左子树，根，右子树

def mid(tree):
    l = []

    def deep(tree):
        if not tree:
            return
        deep(tree.left)
        l.append(tree.data)
        deep(tree.right)
    deep(tree)
    return l

后序遍历：左子树，右子树，根

def post(tree):
    l = []

    def deep(tree):
        if not tree:
            return
        deep(tree.left)
        deep(tree.right)
        l.append(tree.data)
    deep(tree)
    return l

3. 已知前序，中序求后序

def rebuild(pre, mid):
    if not pre:
        return
    data = pre[0]
    tree = Node(data)
    index = mid.index(data)
    tree.left = rebuild(pre[1:index+1], mid[:index])
    tree.right = rebuild(pre[index+1:], mid[index+1:])
    return tree

def post(tree):
    l = []

    def deep(tree):
        if not tree:
            return
        deep(tree.left)
        deep(tree.right)
        l.append(tree.data)
    deep(tree)
    return l

分析：

前序：[1, 3, 7, 0, 6, 2, 5]

中序：[0, 7, 3, 6, 1, 5, 2]

得知：

1、 根节点为1

2、 3, 7, 0, 6为左子树的前序

3、 0, 7, 3, 6为左子树的中序

4、2, 5为右子树的前序

5、5, 2为右子树的中序

根据2和3依此规律可构建一棵左子树

根据4和5依此规律可构建一棵右子树

4. 二叉树的最大深度

def max_depth(tree):
    if not tree:
        return 0
    return max(max_depth(tree.left), max_depth(tree.right)) + 1

5. 判断两棵二叉树是否相同

def is_same_tree(tree1, tree2):
    if tree1 and tree2:
        return tree1.data == tree2.data and is_same_tree(tree1.left, tree2.left) and is_same_tree(tree1.right, tree2.right)
    elif not tree1 and not tree2:
        return True
    else:
        return False

6. 序列化与反序列化

二叉树的序列化是指：把一棵二叉树按照某种遍历方式的结果以某种格式保存为字符串，从而使得内存中建立起来的二叉树可以持久保存。

某种遍历方式是指 可以基于 按层、 先序、中序、后序 的二叉树遍历方式

某种格式是指 用某种符号表示一个空节点（如#），用某种符号表示结点值的结束（如！）

def serialize(tree):
    l = []

    def deep(tree):
        if not tree:
            l.append('#!')
            return
        l.append('%s!' % tree.data)
        deep(tree.left)
        deep(tree.right)
    deep(tree)
    return ''.join(l)

def deserialize(str):
    l = str.split('!')

    def deep(l):
        if l[0] == '#':
            l.pop(0)
            return
        tree = Node(int(l.pop(0)))
        tree.left = deep(l)
        tree.right = deep(l)
        return tree
    return deep(l)

7. 判断二叉树是否是平衡二叉树

 

单链表

1. 建立链表

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def createLinkList(l):
    if not l:
        return
    head = Node(l.pop(0))
    cur = head
    for data in l:
        node = Node(data)
        cur.next = node
        cur = cur.next
    return head

2. 遍历链表

def travel(head):
    cur = head
    while cur:
        print(cur.data)
        cur = cur.next

3. 单链表逆置

def rev(head):
    if not head:
        return
    pre = head
    cur = head.next
    pre.next = None
    while cur:
        temp = cur.next
        cur.next = pre
        pre = cur
        cur = temp
    return pre

4. 合并两个有序链表

def mergeLinkList(head1, head2):
    if not head1:
        return head2
    if not head2:
        return head1
    head = None
    if head1.data < head2.data:
        head = Node(head1.data)
        head1 = head1.next
    else:
        head = Node(head2.data)
        head2 = head2.next
    head.next = mergeLinkList(head1, head2)
    return head