Python 递归式实现二叉树前序,中序,后序遍历
时间:2022-06-10 08:27:35|栏目:Python代码|点击: 次
记忆点:
- 前序:VLR
- 中序:LVR
- 后序:LRV
举例:
一颗二叉树如下图所示:
则它的前序、中序、后序遍历流程如下图所示:
1.前序遍历
class Solution: def preorderTraversal(self, root: TreeNode): def preorder(root: TreeNode): if not root: return res.append(root.val) preorder(root.left) preorder(root.right) res = [] preorder(root) return res
2.中序遍历
class Solution: def inorderTraversal(self, root: TreeNode): def inorder(root: TreeNode): if not root: return inorder(root.left) res.append(root.val) inorder(root.right) res = [] inorder(root) return res
3.后序遍历
class Solution: def postorderTraversal(self, root: TreeNode): def postorder(root: TreeNode): if not root: return postorder(root.left) res.append(root.val) postorder(root.right) res = [] postorder(root) return res
4.测试
class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right # 用列表递归创建二叉树 def createTree(root,list_n,i): if i <len(list_n): if list_n[i] == 'null': return None else: root = TreeNode(val = list_n[i]) root.left = createTree(root.left,list_n,2*i+1) root.right = createTree(root.right,list_n,2*i+2) return root return root class Solution: def preorderTraversal(self, root: TreeNode): # 前序 def preorder(root: TreeNode): if not root: return res.append(root.val) preorder(root.left) preorder(root.right) res = [] preorder(root) return res def inorderTraversal(self, root: TreeNode): # 中序 def inorder(root: TreeNode): if not root: return inorder(root.left) res.append(root.val) inorder(root.right) res = [] inorder(root) return res def postorderTraversal(self, root: TreeNode): # 后序 def postorder(root: TreeNode): if not root: return postorder(root.left) postorder(root.right) res.append(root.val) res = [] postorder(root) return res if __name__ == "__main__": root = TreeNode() list_n = [1,2,3,4,5,6,7,8,'null',9,10] root = createTree(root,list_n,0) s = Solution() res_pre = s.preorderTraversal(root) res_in = s.inorderTraversal(root) res_post = s.postorderTraversal(root) print(res_pre) print(res_in) print(res_post)
5.结果
6.补充
6.1N叉树前序遍历
""" # Definition for a Node. class Node: def __init__(self, val=None, children=None): self.val = val self.children = children """ class Solution: def postorder(self, root: 'Node') -> List[int]: def seq(root): if not root: return res.append(root.val) for child in root.children: seq(child) res = [] seq(root) return res N叉树后序遍历 """ # Definition for a Node. class Node: def __init__(self, val=None, children=None): self.val = val self.children = children """ class Solution: def postorder(self, root: 'Node') -> List[int]: def seq(root): if not root: return for child in root.children: seq(child) res.append(root.val) res = [] seq(root) return res
栏 目:Python代码
下一篇:如何使用python-dotenv解决代码与敏感信息的分离
本文标题:Python 递归式实现二叉树前序,中序,后序遍历
本文地址:http://www.codeinn.net/misctech/204369.html