Leetcode刷题笔记1 二叉树part01

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#leetcode #笔记 #算法

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二叉树的递归遍历

leetcode 144 二叉树的前序遍历

class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res = []
        def dfs(node):
            if node is None:
                return 
            res.append(node.val)
            dfs(node.left)
            dfs(node.right)
        dfs(root)
        return res

leetcode 94 二叉树的中序遍历

class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res = []
        def dfs(node):
            if node is None:
                return
            dfs(node.left)
            res.append(node.val)
            dfs(node.right)
        dfs(root)
        return res

leetcode 145 二叉树的后序遍历

class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res = []
        def dfs(node):
            if node is None:
                return
            dfs(node.left)
            dfs(node.right)
            res.append(node.val)
        dfs(root)
        return res

二叉树的迭代遍历

中序遍历

class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        res = []
        stack = []
        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

前序遍历

class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        stack = [root]
        res = []
        while stack:
            node = stack.pop()
            res.append(node.val)
            if node.right:
                stack.append(node.right)
            if node.left:
                stack.append(node.left)
        return res

后序遍历

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]

leetcode 102 二叉树的层序遍历

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        queue = collections.deque([root])
        res = []
        while queue:
            level = []
            for _ in range(len(queue)):
                cur = queue.popleft()
                level.append(cur.val)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            res.append(level)
        return res

递归

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        levels = []
        def traverse(node, level):
            if not node:
                return
            if len(levels) == level:
                levels.append([])
            levels[level].append(node.val)
            traverse(node.left, level + 1)
            traverse(node.right, level + 1)
        traverse(root, 0)
        return levels