标签:turn def ati int 情况 find 删除 lis and
class TreeNode: def __init__(self, val): self.val = val self.left = None self.right = None class OperationTree: def insert(self, root, val): if root is None: root = TreeNode(val) elif val < root.val: root.left = self.insert(root.left, val) elif val > root.val: root.right = self.insert(root.right, val) return root def printTree(self, root): if root is None: return self.printTree(root.left) print(root.val, end=‘ ‘) self.printTree(root.right) def query(self, root): if root is None: return if val == root.val: return True elif val < root.val: self.query(root.left) elif val > root.right: self.query(root.right) def find_min(self, root): if root.left: return self.find_min(root.left) else: return root def find_max(self, root): if root.right: return self.find_max(root.right) else: return root def del_node(self, root, val): if root is None: return if val < root.val: root.left = self.del_node(root.left, val) elif val > root.val: root.right = self.del_node(root.right, val) # 当val == root.val时,分为三种情况:只有左子树或者只有右子树、有左右子树、即无左子树又无右子树 else: if root.left and root.right: # 既有左子树又有右子树,则需找到右子树中最小值节点 temp = self.find_min(root.right) root.val = temp.val # 再把右子树中最小值节点删除,让它符合二叉搜索树的定义 root.right = self.del_node(root.right, temp.val) elif root.right is None and root.left is None: # 左右子树都为空 root = None elif root.right is None: # 只有左子树 root = root.left elif root.left is None: # 只有右子树 root = root.right return root if __name__ == ‘__main__‘: List = [17, 5, 35, 2, 11, 29, 38, 9, 16, 8] root = None op = OperationTree() for val in List: root = op.insert(root, val) print(‘中序打印二叉搜索树:‘, end=‘ ‘) op.printTree(root)
标签:turn def ati int 情况 find 删除 lis and
原文地址:https://www.cnblogs.com/c-x-a/p/11065991.html