排序二叉树的实现

Posted 2021-02-20 sun-10387834

# coding:utf8
class Node():
    def __init__(self, _item):
        self.item = _item
        self.left = None
        self.right = None


class SortTree():
    def __init__(self):
        self.root = None

    def add(self, _item):
        node = Node(_item)
        cur = self.root
        # 如果插入的第一个元素为根节点
        if cur == None:
            self.root = node
            return
        # 如果插入元素是叶子节点
        while True:
            ##往左插入
            if _item < cur.item:
                if cur.left == None:
                    cur.left = node
                    break
                else:
                    cur = cur.left
            ##往右插入
            else:
                if cur.right == None:
                    cur.right = node
                    break
                else:
                    cur = cur.right

    # 深度优先搜索遍历
    def forward(self, root):  # 前序遍历 ：根左右
        if root == None:
            return
        print(root.item)
        self.forward(root.left)
        self.forward(root.right)

    def middle(self, root):  # 中序遍历 ：左根右
        if root == None:
            return
        self.middle(root.left)
        print(root.item)
        self.middle(root.right)

    def back(self, root):  # 后序遍历 ：左右根
        if root == None:
            return
        self.back(root.left)
        self.back(root.right)
        print(root.item)


if __name__ == "__main__":
    tree = SortTree()
    alist = [3, 8, 5, 7, 6, 2, 9, 4, 1]
    for i in alist:
        tree.add(i)
    tree.forward(tree.root)
    tree.middle(tree.root)
    tree.back(tree.root)

由打印结果可得：中序遍历得到的是排序后的结果。