力扣python实现-二叉树合集

• 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

• 144.二叉树的前序遍历

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

• 145.二叉树的后序遍历

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

• 94.二叉树的中序遍历

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

• 102.二叉树的层序遍历
• 103.二叉树的锯齿形层序遍历

• 104.二叉树的最大深度

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

• 111.二叉树的最小深度

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

• 101.对称二叉树

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

• 105.从前序与中序遍历序列构造二叉树

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   	

• 106.从中序与后序遍历序列构造二叉树

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

• 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

• 236.二叉树的最近公共祖先
• 235.二叉搜索树的最近公共祖先

• 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