文章目录
- 二叉树的遍历方式
- 二叉树的属性
- 二叉树的修改与构造
- 求二叉搜索树的属性
- 二叉树公共祖先问题
- 二叉搜索树的修改与构造
二叉树的遍历方式
class Solution:
def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
res = []
def dfs(root):
if root is None:
return
res.append(root.val)
dfs(root.left)
dfs(root.right)
dfs(root)
return res
- 用栈进行前序遍历，顺序是根右左
def preorderTraversal(self,root):
res = []
if root is None:
return res
stack = [root]
while stack:
pop_node = stack.pop()
res.append(pop_node.val)
if pop_node.right is not None:
stack.append(pop_node.right)
if pop_node.left is not None:
stack.append(pop_node.left)
return res
class Solution:
def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
res = []
def dfs(root):
if root is None:
return
dfs(root.left)
dfs(root.right)
res.append(root.val)
dfs(root)
return res
class Solution:
def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
res = []
stack = [root]
while stack:
pop_node = stack.pop()
res.append(pop_node.val)
if pop_node.left is not None:
stack.append(pop_node.right)
if pop_node.right is not None:
stack.append(pop_node.left)
return res[::-1]
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
res = []
def dfs(root):
if root is None:
return
dfs(root.left)
res.append(root.val)
dfs(root.right)
dfs(root)
return res
- 需要借助一个指针，并且一上来不需要把self.root先放到stack里.
- 先找到树的左边的节点，依次压入占中，直到找到最左边的节点，cur循环结束
- 将最左边的节弹出后，然后加入res
- 若当前节点存在右节点，则令cur等于右节点
def inorder_stack(self, root:TreeNode):
res，stack = [], []
if root is None:
return res
cur = root
while stack or cur:
while cur:
stack.append(cur)
cur = cur.left
pop_node = stack.pop()
res.append(pop_node)
cur = pop_node.right
return res
- 102.二叉树的层序遍历
- 103.二叉树的锯齿形层序遍历
二叉树的属性
class Solution:
def maxDepth(self, root: TreeNode) -> int:
if root is None:
return 0
return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1
class Solution:
def minDepth(self, root: TreeNode) -> int:
if root is None:
return 0
elif root.left is not None and root.right is not None:
return min(self.minDepth(root.left), self.minDepth(root.right)) + 1
else:
return max(self.minDepth(root.left), self.minDepth(root.right)) + 1
class Solution:
def isSymmetric(self, root: Optional[TreeNode]) -> bool:
def isSym(left, right):
if left and right:
return left.val == right.val and isSym(left.left, right.right) and isSym(left.right, right.left)
else:
return left == right
return isSym(root, root)
- 110.平衡二叉树
- 222.完全二叉树的节点个数
- 257.二叉树的所有路径
- 404.左叶子之和
- 513.找树左下角的值
- 112.路径之和
二叉树的修改与构造
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
if not preorder and not inorder:
return None
root = TreeNode(preorder[0])
ind = inorder.index(root)
left_node = self.buildTree(preorder[1:ind], inorder[ind+1:])
right_node = self.buildTree(preorder[ind+1:], inorder[ind+1:])
root.left = left_node
root.right = right_node
return root
class Solution:
def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
if not inorder and not postorder:
return None
root = TreeNode(postorder[-1])
ind = inorder.index(root)
left_node = self.buildTree(inorder[:ind], postorder[:ind])
right_node = self.buildTree(inorder[ind+1:], postorder[ind:-1])
root.left = left_node
root.right = right_node
return root
- 226.翻转二叉树
- 654.最大二叉树
- 617.合并二叉树
求二叉搜索树的属性
- 98.验证二叉搜索树
- 700.二叉搜索树中的搜索
- 530.二叉搜索树的最小绝对差
- 501.二叉搜索树中的众数
- 538.把二叉搜索树转换为累加树
- 230.二叉搜索树中第K小的元素
二叉树公共祖先问题
- 236.二叉树的最近公共祖先
- 235.二叉搜索树的最近公共祖先
二叉搜索树的修改与构造
- 701.二叉搜索树中的插入操作
- 450.删除二叉搜索树中的节点
- 669.修剪二叉搜索树
- 108.将有序数组转换为二叉搜索树
