Brush the topic-BinaryTree
大家好，这是Brush the topic的第一章节,BinaryTree。首先我说一下为什么把这个放在刷题的第一节呢？
原因如下:
- 培养、训练自己的计算机的思维。
- 锻炼模版化，抽象化思维
下面让我们一起去完成一个壮举，那就是完全解决二叉树的遍历问题，以及相关问题。are you ok？
知识点回顾
二叉树的遍历
由于对于二叉树的遍历顺序不同，构造出三种不同的遍历方式
- 前序遍历-根左右
- 中序遍历-左根右
- 后序遍历-左右根
递归代码模版如下
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
# 前序遍历
def preOreder(self, root):
if root:
self.traverse_path.append(root.val)
preOreder(self.left)
preOreder(self.right)
# 中序遍历
def inOreder(self, root):
if root:
preOreder(self.left)
self.traverse_path.append(root.val)
preOreder(self.right)
# 后序遍历
def postOreder(self, root):
if root:
preOreder(self.left)
preOreder(self.right)
self.traverse_path.append(root.val)
Golang
// 前序遍历
func preOreder(root *TreeNode) {
result := []int{
}
if root == nil {
return}
result = append(result, root.value)
preOreder(root.Left)
preOreder(root.Right)
}
// 中序遍历
func inOreder(root *TreeNode) {
result := []int{
}
if root == nil {
return}
preOreder(root.Left)
result = append(result, root.value)
preOreder(root.Right)
}
// 后序遍历
func postOreder(root *TreeNode) {
result := []int{
}
if root == nil {
return}
postOreder(root.Left)
postOreder(root.Right)
result = append(result, root.value)
}
practice
基于此我们可以拿下以下题目,完全二叉树递归模版解题
144. 二叉树的前序遍历-Python
Recursive
# 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
# Recursive-1
class Solution:
def preorderTraversal(self, root: TreeNode) -> List[int]:
result = []
self.helper(root, result)
return result
def helper(self, root, result):
if root is None: return
result.append(root.val)
self.helper(root.left,result)
self.helper(root.right, result)
# Recursive-2 Another way Anonymous function
class Solution:
def preorderTraversal(self, root: TreeNode) -> List[int]:
def helper(root: TreeNode):
if not root: return
res.append(root.val)
helper(root.left)
helper(root.right)
res = list()
helper(root)
return res
# Recursive-3 more clean code
class Solution:
def preorderTraversal(self, root: TreeNode) -> List[int]:
if not root:return []
res = []
res.append(root.val)
res+=self.preorderTraversal(root.left)
res+=self.preorderTraversal(root.right)
return res
Iterative
# 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
# Solution-1
class Solution1:
def preorderTraversal(self, root: TreeNode) -> List[int]:
stack, result = [], []
while stack or root:
while root:
# 前序遍历-根左右，先拿根
result.append(root.val)
# 压栈
stack.append(root)
# 拿完根之后拿左儿子
root = root.left
# 左儿子拿出来，拿右儿子
node = stack.pop()
root = node.right
# # 完成
return result
# Solution-2 简化Solution-1
class Solution2:
def preorderTraversal(self, root: TreeNode) -> List[int]:
stack, result = [], []
while stack or root:
if root:
result.append(root.val)
stack.append(root)
root = root.left
else:
node = stack.pop()
root = node.right
return result
# Solution-3
class Solution3:
def preorderTraversal(self, root: TreeNode) -> List[int]:
stack, result = [root], []
while stack:
# 拿出根
node = stack.pop()
if node:
# 前序遍历拿出，先拿根的值
result.append(node.val)
# 模仿栈，先入后出。后拿右孩子
stack.append(node.right)
stack.append(node.left)
return result
94. 二叉树的中序遍历-Python
Recursive
# 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
# Recursive-1
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
result = []
self.helper(root, result)
return result
def helper(self, root, result):
if root is None: return
self.helper(root.left,result)
result.append(root.val)
self.helper(root.right, result)
# Recursive-2 Another way Anonymous function
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
def helper(root: TreeNode):
if not root: return
helper(root.left)
res.append(root.val)
helper(root.right)
res = list()
helper(root)
return res
# Recursive-3 more clean code
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
if not root:return []
res = []
res+=self.preorderTraversal(root.left)
res.append(root.val)
res+=self.preorderTraversal(root.right)
return res
Iterative
# 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
# Solution - 1
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
if not root: return
stack, result = [], []
while stack or root:
while root:
stack.append(root)
root = root.left
node = stack.pop()
result.append(node.val)
root = node.right
return result
# Solution - 2 简化Solution-1
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
stack, result = [], []
while stack or root:
if root:
stack.append(root)
root = root.left
else:
node = stack.pop()
result.append(node.val)
root = node.right
return result
# Solution - 3
class Solution2:
def inorderTraversal(self, root: TreeNode) -> List[int]:
stack, result = [], []
while stack or root:
if root:
stack.append(root)
root = root.left
else:
node = stack.pop()
result.append(node.val)
root = node.right
return result
145. 二叉树的后序遍历
Recursive
# 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
# Recursive-1
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
result = []
self.helper(root, result)
return result
def helper(self, root, result):
if root is None: return
self.helper(root.left,result)
self.helper(root.right, result)
result.append(root.val)
# Recursive-2 Another way Anonymous function
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
def helper(root: TreeNode):
if not root: return
helper(root.left)
helper(root.right)
res.append(root.val)
res = list()
helper(root)
return res
# Recursive-3 more clean code
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
if not root:return []
res = []
res+=self.preorderTraversal(root.left)
res+=self.preorderTraversal(root.right)
res.append(root.val)
return res
Iterative
# Solution - 1
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
stack, result = [], []
while root or stack:
while root:
result.append(root.val)
stack.append(root)
root = root.right
node = stack.pop()
root = node.left
return result[::-1]
# Solution - 2
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
stack, result = [], []
while stack or root:
if root:
result.append(root.val)
stack.append(root)
root = root.right
else:
node = stack.pop()
root = node.left
return result[::-1]
# Solution - 3
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
if not root: return
stack, result = [root], []
while stack:
node = stack.pop()
if node:
result.append(node.val)
stack.append(node.left)
stack.append(node.right)
return result[::-1]
二叉树迭代遍历模版-Python
# 前序遍历
# Solution-1
class Solution1:
def preorderTraversal(self, root: TreeNode) -> List[int]:
if not root: return
stack, result = [], []
while root or stack:
while root:
result.append(root.val)
stack.append(root)
root = root.left
tmp = stack.pop()
root = tmp.right
return result
# 中序遍历
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
if not root: return
stack, result = [], []
while stack or root:
while root:
stack.append(root)
root = root.left
node = stack.pop()
result.append(node.val)
root = node.right
return result
由递归到迭代，基本的思想就是由递归中由系统维护的栈，转为手动维护。