Python实现二叉树结构及三种遍历 教你怎样使用Python实现二叉树结构及三种遍历

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)

# 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
    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]
"""