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