《数据结构与算法分析python版》第九周编程作业
《数据结构与算法分析python版》第九周编程作业
1、二叉树复原
题目内容：
给定一种序列化二叉树的方式：从根节点起始按层次遍历二叉树所有“可能”存在节点的位置：若该位置存在节点，则输出节点值，并在下一层相应增加两个可用位置；否则输出None，且不增加下一层的可用位置。
例如"[5, 4, 7, 3, None, 2, None, -1, None, 9]"是下图所示的二叉树序列化的结果：
其中红色箭头对所有的None进行了标记。
现给出一个二叉树以这种形式序列化的结果，请复原该二叉树并给出它的中序遍历。
输入格式:
一行合法的Python表达式，可解析为包含整数与None的列表
输出格式：
二叉树中序遍历的整数序列，以空格分隔
若无对应下标，则输出"none"
输入样例
[5, 4, 7, 3, None, 2, None, -1, None, 9]
输出样例：
-1 3 4 5 9 2 7
没得思路，转一下别人的代码
脚本
class BinaryTree:
def __init__(self,data):
self.data = data
self.leftChild = None
self.rightChild = None
self.leftflag = 0
self.rightflag = 0
def seq2tree(seq):
queue = []
i = 1
currentTree = BinaryTree(seq[0])
wholeTree = currentTree
queue.append(currentTree)
while i < len(seq) and len(queue)> 0:
currentTree = queue.pop(0)
if seq[i] == None:
if currentTree.leftflag == 0 and currentTree.rightflag == 0:
currentTree.leftflag = 1
queue.insert(0, currentTree)
elif currentTree.leftflag == 1 and currentTree.rightflag == 0:
currentTree.rightflag = 1
i = i + 1
else:
if currentTree.leftflag == 0 and currentTree.rightflag == 0:
currentTree.leftChild = BinaryTree(seq[i])
currentTree.leftflag = 1
queue.append(currentTree.leftChild)
queue.insert(0,currentTree)
i = i +1
elif currentTree.leftflag == 1 and currentTree.rightflag == 0:
currentTree.rightChild = BinaryTree(seq[i])
currentTree.rightflag = 1
queue.append(currentTree.rightChild)
i = i +1
else:
i = i
return wholeTree
def inorderTree(root):
if root == None:
return []
else:
left = inorderTree(root.leftChild)
data = root.data
right = inorderTree(root.rightChild)
return left + [data] + right
lst = eval(input())
tree = seq2tree(lst)
inorder = inorderTree(tree)
print(' '.join(str(x) for x in inorder))
2、翻转二叉树
题目内容：
给定一个二叉树，请给出它的镜面翻转。
为方便起见，本题只给出完全二叉树的层次遍历，请给出相应的翻转二叉树的中序遍历。
输入格式:
一行空格分隔的整数序列，表示一个完全二叉树的层次遍历
输出格式：
一行空格分隔的整数序列，表示翻转后的二叉树的中序遍历
输入样例
4 2 7 1 3 6 9
输出样例：
9 7 6 4 3 2 1
脚本
这道题接着上面的就容易多了
class BinaryTree:
def __init__(self,data):
self.data = data
self.leftChild = None
self.rightChild = None
self.leftflag = 0
self.rightflag = 0
def seq2tree(seq):
queue = []
i = 1
currentTree = BinaryTree(seq[0])
wholeTree = currentTree
queue.append(currentTree)
while i < len(seq) and len(queue)> 0:
currentTree = queue.pop(0)
if currentTree.leftflag == 0 and currentTree.rightflag == 0:
currentTree.leftChild = BinaryTree(seq[i])
currentTree.leftflag = 1
queue.append(currentTree.leftChild)
queue.insert(0,currentTree)
i = i +1
elif currentTree.leftflag == 1 and currentTree.rightflag == 0:
currentTree.rightChild = BinaryTree(seq[i])
currentTree.rightflag = 1
queue.append(currentTree.rightChild)
i = i +1
else:
i = i
return wholeTree
def invertTree(wholeTree):
if not wholeTree:
return []
wholeTree.leftChild, wholeTree.rightChild = wholeTree.rightChild, wholeTree.leftChild
invertTree(wholeTree.leftChild)
invertTree(wholeTree.rightChild)
return wholeTree
def inorderTree(root):
if root == None:
return []
else:
left = inorderTree(root.leftChild)
data = root.data
right = inorderTree(root.rightChild)
return left + [data] + right
lst = list(input().split(" "))
tree = seq2tree(lst)
invertedTree = invertTree(tree)
inorder = inorderTree(invertedTree)
print(' '.join(str(x) for x in inorder))
3、多叉树遍历
题目内容：
给定以嵌套列表形式给出的多叉树，求它的后序遍历
注：每个代表非空多叉树的列表包含至少一项；列表第一项代表节点值，其后每一项分别为子树；遍历子树时以列表下标从小到大的顺序进行
输入格式:
一行合法的Python表达式，可解析为嵌套列表形式的多叉树结构
输出格式：
一行整数，以空格分隔
输入样例
[1,[2,[3,[4],[5]],[6]],[7],[8,[9],[10]]]
输出样例：
4 5 3 6 2 7 9 10 8 1
脚本
代码
def postorderTree(tree):
if len(tree) == 1:
return tree
else:
result = []
for i in range(1, len(tree)):
result += postorderTree(tree[i])
result += [tree[0]]
return result
lst = eval(input())
postorder = postorderTree(lst)
print(*postorder)