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25期代码随想录算法训练营第十四天 | 二叉树 | 递归遍历、迭代遍历
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目录
- 递归遍历
- 前序遍历
- 中序遍历
- 后序遍历
- 迭代遍历
- 前序遍历
- 中序遍历
- 后序遍历
递归遍历
前序遍历
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:if not root:return []left = self.preorderTraversal(root.left)right = self.preorderTraversal(root.right)return [root.val] + left + right
中序遍历
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:if not root:return []left = self.inorderTraversal(root.left)right = self.inorderTraversal(root.right)return left + [root.val] + right
后序遍历
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:if not root:return []left = self.postorderTraversal(root.left)right = self.postorderTraversal(root.right)return left + right + [root.val]
迭代遍历
前序遍历
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:if not root:return []stack = [root]res = []while stack:cur = stack.pop()res.append(cur.val) # 在这里加入节点值# 先右后左地加入子节点到栈中if cur.right:stack.append(cur.right)if cur.left:stack.append(cur.left)return res
中序遍历
为什么这样能实现左中右的逻辑?
为什么需要使用while cur or stack
?
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:if not root:return []stack = []res = []cur = rootwhile cur or stack:if cur:stack.append(cur)cur = cur.leftelse:cur = stack.pop()res.append(cur.val)cur = cur.rightreturn res
后序遍历
中右左 ---->再颠倒顺序成为 左右中
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:if not root:return []res = []stack = [root]while stack:node = stack.pop()res.append(node.val)if node.left:stack.append(node.left)if node.right:stack.append(node.right)return res[::-1]
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