剑指offer-顺序打印二叉树节点（系列）-树-python

转载自  https://blog.csdn.net/u010005281/article/details/79761056 非常感谢！

首先创建二叉树，然后按各种方式打印：

class treeNode:
    def __init__(self, x):
        self.left = None
        self.right = None
        self.val = x

class Solution:

    # 给定二叉树的前序遍历和中序遍历，获得该二叉树
    def getBSTwithPreTin(self, pre, tin):
        if not pre or not tin:
            return None
        root = treeNode(pre[0])
        for order, item in enumerate(tin):
            if root.val == item:
                root.left = self.getBSTwithPreTin(pre[1:order + 1], tin[:order])
                root.right = self.getBSTwithPreTin(pre[order + 1:], tin[order + 1:])
                return root

   #从上往下打印出二叉树的每个节点，同层节点从左至右打印
    #使用队列先进先出来依次打印
    def PrintFromTopToBottom(self, root):
        array = []
        result = []
        if root == None:
            return result
        #root为根，其值为根节点的值
        array.append(root)
        while array:
            newNode = array.pop(0)
            result.append(newNode.val)
            if newNode.left != None:
                array.append(newNode.left)#先放左边
            if newNode.right != None:
                array.append(newNode.right)
        return result

    #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。
    def Print(self, pRoot):
        if not pRoot:
            return []
        resultList = []
        curLayer = [pRoot]
        count = 0
        while curLayer:
            curList = []
            nextLayer = []
            for node in curLayer:
                curList.append(node.val)
                if node.left:
                    nextLayer.append(node.left)
                if node.right:
                    nextLayer.append(node.right)
            if count // 2 == 0:
                curList.reverse()
            resultList.append(curList)
            curLayer = nextLayer

        return resultList

    #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。

    #方式1：
    #1. 按序获取每一层节点的值；
    # 2. 将偶数层节点的值倒序。
    def PrintZ_1(self, pRoot):
        # write code here
        if pRoot == None:
            return []
        cur_layer = [pRoot]
        res = []
        isEvenLayer = True
        while cur_layer:
            curlist = []
            nextlayer = []
            isEvenLayer = not isEvenLayer #是否倒叙
            for node in cur_layer:
                curlist.append(node.val)
                if node.left:
                    nextlayer.append(node.left)
                if node.right:
                    nextlayer.append(node.right)
            if isEvenLayer == False:
                res.append(curlist)
            else:
                res.append(curlist[::-1])
            cur_layer = nextlayer
        return res

    #方式2：
    # 获取每一层的节点的值时，如果是偶数层，则将每个节点的值插入到列表的头部，
    # 即实现了获取节点值时如果是偶数层则倒序排列的效果：
    def PrintZ_2(self, pRoot):
        # write code here
        if pRoot == None:
            return []
        cur_layer = [pRoot]
        res = []
        isEvenLayer = True
        while cur_layer:
            curlist = []
            nextlayer = []
            isEvenLayer = not isEvenLayer  # 是否倒叙
            for node in cur_layer:
                if isEvenLayer == False:
                    curlist.append(node.val)
                else:
                    curlist.insert(0,node.val)

                if node.left:
                    nextlayer.append(node.left)
                if node.right:
                    nextlayer.append(node.right)
            res.append(curlist)
            cur_layer = nextlayer
        return res

if __name__ == '__main__':
    # flag = "printTreeNode"
    # flag = "printTreeNode_line"
    # flag = "printTreeNode_Z1"
    flag = "printTreeNode_Z2"
    solution = Solution()
    preorder_seq = [1, 2, 4, 7, 3, 5, 6, 8]
    middleorder_seq = [4, 7, 2, 1, 5, 3, 8, 6]
    treeRoot1 = solution.getBSTwithPreTin(preorder_seq, middleorder_seq)
    if flag == "printTreeNode":
        newArray = solution.PrintFromTopToBottom(treeRoot1)
        print(newArray)
    elif flag == "printTreeNode_line":
        newArray = solution.Print(treeRoot1)
        print(newArray)
    elif flag == "printTreeNode_Z1":
        newArray = solution.PrintZ_1(treeRoot1)
        print(newArray)
    else:
        newArray = solution.PrintZ_2(treeRoot1)
        print(newArray)