教你如何使用Python实现二叉树结构及三种遍历

时间:2022-04-02 07:35:26

一:代码实现

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class TreeNode:
    """节点类"""
    def __init__(self, mid, left=None, right=None):
        self.mid = mid
        self.left = left
        self.right = right
 
 
# 树类
class Tree:
    """树类"""
    def __init__(self, root=None):
        self.root = root
 
    def add(self, item):
        # 将要添加的数据封装成一个node结点
        node = TreeNode(item)
        if not self.root:
            self.root = node
            return
        queue = [self.root]
        while queue:
            cur = queue.pop(0)
            if not cur.left:
                cur.left = node
                return
            else:
                queue.append(cur.left)
 
            if not cur.right:
                cur.right = node
                return
            else:
                queue.append(cur.right)
               
tree = Tree()
tree.add(0)
tree.add(1)
tree.add(2)
tree.add(3)
tree.add(4)
tree.add(5)
tree.add(6)

二:遍历

在上述树类代码基础上加遍历函数,基于递归实现。

教你如何使用Python实现二叉树结构及三种遍历

先序遍历:

先序遍历结果是:0 -> 1 -> 3 -> 4 -> 2 -> 5 -> 6

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# 先序遍历
    def preorder(self, root, result=[]):
        if not root:
            return
        result.append(root.mid)
        self.preorder(root.left, result)
        self.preorder(root.right, result)
        return result
        
print("先序遍历")
print(tree.preorder(tree.root))
"""
先序遍历
[0, 1, 3, 4, 2, 5, 6]
"""

中序遍历:

中序遍历结果是:3 -> 1 -> 4 -> 0 -> 5 -> 2 -> 6

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# 中序遍历
    def inorder(self, root, result=[]):
        if not root:
            return result
        self.inorder(root.left, result)
        result.append(root.mid)
        self.inorder(root.right, result)
        return result
        
print("中序遍历")
print(tree.inorder(tree.root))
"""
中序遍历
3, 1, 4, 0, 5, 2, 6]
"""

后续遍历

后序遍历结果是:3 -> 4 -> 1 -> 5 -> 6 -> 2 -> 0

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# 后序遍历
    def postorder(self, root, result=[]):
        if not root:
            return result
        self.postorder(root.left, result)
        self.postorder(root.right, result)
        result.append(root.mid)
 
        return result
        
print("后序遍历")
print(tree.postorder(tree.root))
"""
后序遍历
[3, 4, 1, 5, 6, 2, 0]
"""

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原文链接:https://blog.csdn.net/qq_39434183/article/details/117927680