剑指offer 面试32题

面试32题：

题目：从上到下打印二叉树

题：不分行从上到下打印二叉树

解题代码：

# -*- coding:utf-8 -*-

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

    # 返回从上到下每个节点值列表，例：[1,2,3]

    def PrintFromTopToBottom(self, root):

        # write code here

        if not root:

            return []

        res=[]

        res_val=[]

        res.append(root)

        while len(res)>0:

            node=res.pop(0)

            res_val.append(node.val)

            if node.left:

                res.append(node.left)

            if node.right:

                res.append(node.right)

        return res_val

题目拓展一：分行从上到下打印二叉树。

题：从上到下按层打印二叉树，同一层结点从左至右输出。每一层输出一行。

解题代码一：同剑指offer

# -*- coding:utf-8 -*-

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

    # 返回二维列表[[1,2],[4,5]]

    def Print(self, pRoot):

        # write code here

        if not pRoot:

            return []

        res=[]

        res_val=[]

        res.append(pRoot)

        nextLevel=0 #表示下一层节点的数目

        toBePrinted=1 #表示当前层还没有打印的节点数

        temp=[]

        while len(res)>0:

            node=res[0]

            temp.append(node.val)

            if node.left:

                res.append(node.left)

                nextLevel+=1

            if node.right:

                res.append(node.right)

                nextLevel+=1

            del res[0]

            toBePrinted-=1

            if toBePrinted==0:

                res_val.append(temp)

                toBePrinted=nextLevel

                nextLevel=0

                temp=[]

        return res_val

解题代码二：代码更简洁。

# -*- coding:utf-8 -*-

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

    # 返回二维列表[[1,2],[4,5]]

    def Print(self, pRoot):

        # write code here

        if not pRoot:

            return []

        res,nodes=[],[pRoot]

        while nodes:

            curStack,nextStack=[],[]

            for node in nodes:

                curStack.append(node.val)

                if node.left:

                    nextStack.append(node.left)

                if node.right:

                    nextStack.append(node.right)

            res.append(curStack)

            nodes=nextStack

        return res

解题代码三：cur、last记录，思路同二

# -*- coding:utf-8 -*-

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

    # 返回二维列表[[1,2],[4,5]]

    def Print(self, pRoot):

        # write code here

        if not pRoot:

            return []

        res=[]

        arr=[]

        arr.append(pRoot)

        cur=0

        #last=1

        while cur<len(arr):

            last=len(arr)

            temp=[]

            while (cur<last):

                temp.append(arr[cur].val)

                if arr[cur].left:

                    arr.append(arr[cur].left)

                if arr[cur].right:

                    arr.append(arr[cur].right)

                cur+=1

            res.append(temp)

        return res

题目拓展二：之字形打印二叉树

题：请实现一个函数按照之字形打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右至左的顺序打印，第三行按照从左到右的顺序打印，其他行以此类推。

解题代码一：简洁。

# -*- coding:utf-8 -*-

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

    def Print(self, pRoot):

        # write code here

        if not pRoot:

            return []

        res=[]

        nodes=[pRoot]

        leftToRight=True

        while nodes:

            curStack,nextStack=[],[]

            for node in nodes:

                curStack.append(node.val)

                if node.left:

                    nextStack.append(node.left)

                if node.right:

                    nextStack.append(node.right)

            if not leftToRight:

                curStack.reverse()

            res.append(curStack)

            leftToRight = not leftToRight

            nodes=nextStack

解题代码二：同剑指offer。

# -*- coding:utf-8 -*-

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

    def Print(self, pRoot):

        # write code here

        if not pRoot:

            return []

        res=[]

        nodes=[pRoot]

        right=True

        while nodes:

            curStack,nextStack=[],[]

            if right:

                for node in nodes:

                    curStack.append(node.val)

                    if node.left:

                        nextStack.append(node.left)

                    if node.right:

                        nextStack.append(node.right)

            else:

                for node in nodes:

                    curStack.append(node.val)

                    if node.right:

                        nextStack.append(node.right)

                    if node.left:

                        nextStack.append(node.left)

            res.append(curStack)

            nextStack.reverse()

            right=not right

            nodes=nextStack

        return res