剑指Offer(Python多种思路实现):从上到下打印二叉树

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

解题思路:

class Solution:
    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。

解题思路一:

# -*- 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

解题思路二:

class Solution:
    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

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。

解题思路一:

# 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

解题思路二:

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
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转载自blog.csdn.net/weixin_44151089/article/details/104471364