目录
递归遍历
前序遍历
python
# 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
python
# 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
python
# 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
python
# 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?
python
# 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 = root
while cur or stack:
if cur:
stack.append(cur)
cur = cur.left
else:
cur = stack.pop()
res.append(cur.val)
cur = cur.right
return res后序遍历
中右左 ---->再颠倒顺序成为 左右中
python
# 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]