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树的遍历

时间:2019-10-06 23:50:14      阅读:71      评论:0      收藏:0      [点我收藏+]

首先是树的建立:

class TreeNode:
    def __init__(self,x,left=None,right=None):
        self.val=x
        self.left=left
        self.right=right
t7=TreeNode(7);t8=TreeNode(8)
t3=TreeNode(3);t4=TreeNode(4,t7,right=None);t5=TreeNode(5,left=None,right=t8);t6=TreeNode(6)
t1=TreeNode(1,t3,t4);t2=TreeNode(2,t5,t6)
root=TreeNode(0,t1,t2)

建立好的树如图所示:

技术分享图片

一、递归版的遍历(很好记)

class traveral:
    def __init__(self):
        self.pre_res=[]
        self.in_res=[]
        self.post_res=[]
    #先序遍历(根左右)
    def preorder(self,root):
        if root is None:
            return None
        self.pre_res.append(root.val)
        self.preorder(root.left)
        self.preorder(root.right)
    #中序遍历(左根右)
    def inorder(self,root):
        if root is None:
            return None
        self.inorder(root.left)
        self.in_res.append(root.val)
        self.inorder(root.right)
    #后序遍历(左右根)
    def postorder(self,root):
        if root is None:
            return None
        self.postorder(root.left)
        self.postorder(root.right)
        self.post_res.append(root.val)

tra=traveral()
tra.preorder(root)
tra.inorder(root)
tra.postorder(root)
print(tra.pre_res)
print(tra.in_res)
print(tra.post_res)

输出:

[0, 1, 3, 4, 7, 2, 5, 8, 6]
[3, 1, 7, 4, 0, 5, 8, 2, 6]
[3, 7, 4, 1, 8, 5, 6, 2, 0]

二、非递归版本

class non_recursive:
    def preorder(self,root):
        stack=[]
        pre_res=[]
        while root or stack:
            while root:
                pre_res.append(root.val)
                stack.append(root)
                root=root.left
            if stack:
                t=stack.pop()
                root=t.right
        return pre_res

    def inorder(self,root):
        stack=[]
        in_res=[]
        while stack or root:
            while root:
                stack.append(root)
                root=root.left
            if stack:
                t=stack.pop()
                in_res.append(t.val)
                root=t.right
        return in_res

    def postorder(self,root):
        stack1=[]
        stack2=[]
        post_res=[]
        stack1.append(root)
        while stack1:
            t=stack1.pop()
            if t.left:
                stack1.append(t.left)
            if t.right:
                stack1.append(t.right)
            stack2.append(t)
        while stack2:
            post_res.append(stack2.pop().val)
        return post_res

    def levelorder(self,root):
        queue=[root]
        level_res=[]
        while queue:
            t=[]
            for i in range(len(queue)):
                cur=queue.pop(0)
                t.append(cur.val)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            level_res.append(t)
        return level_res
nonre=non_recursive()
print(nonre.preorder(root))
print(nonre.inorder(root))
print(nonre.postorder(root))
print(nonre.levelorder(root))

输出:

[0, 1, 3, 4, 7, 2, 5, 8, 6]
[3, 1, 7, 4, 0, 5, 8, 2, 6]
[3, 7, 4, 1, 8, 5, 6, 2, 0]
[[0], [1, 2], [3, 4, 5, 6], [7, 8]]

 

树的遍历

原文:https://www.cnblogs.com/xiximayou/p/11628898.html

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