Leetcode学习day1(二叉树遍历）

python3 参考这篇题解：二叉树遍历https://leetcode-cn.com/problems/binary-tree-preorder-traversal/solution/tu-jie-er-cha-shu-de-si-chong-bian-li-by-z1m/

前序遍历：根->左->右

LeetCode 144.二叉树的前序遍历

迭代解法：（借助栈）

class Solution:

    def preorderTraversal(self, root: TreeNode) -> List[int]:

        if not root: return []

        stack, res=[root], []

        while stack:

            node=stack.pop()

            if node:

                res.append(node.val)

                if node.right:

                    stack.append(node.right)

                if node.left:

                    stack.append(node.left)

        return res

递归解法：

class Solution:

    def preorderTraversal(self, root: TreeNode) -> List[int]:

        res=[]

        def preorder(root):

            nonlocal res

            if not root:

                return

            res.append(root.val)

            preorder(root.left)

            preorder(root.right)

        preorder(root)

        return res

---

中序遍历：左->根->右

Leetcode 94. 二叉树的中序遍历

递归解法：

class Solution:

    def inorderTraversal(self, root: TreeNode) -> List[int]:

        if not root: return []

        stack, res= [(0,root)], []

        while stack:

            flag, node=stack.pop()

            if not node: continue

            if flag==1: res.append(node.val)

            else:

                stack.append((0,node.right))

                stack.append((1,node))

                stack.append((0,node.left))

        return res

递归解法：

class Solution:

    def inorderTraversal(self, root: TreeNode) -> List[int]:

        res=[]

        def inorder(root):

            nonlocal res

            if not root:

               return

            inorder(root.left)

            res.append(root.val)

            inorder(root.right)

        inorder(root)

        return res

---

后序遍历：左->右->根

Leetcode 145. 二叉树的后序遍历

迭代解法：

class Solution:

    def postorderTraversal(self, root: TreeNode) -> List[int]:

        if not root: return []

        stack, res= [(0,root)], []

        while stack:

            flag, node=stack.pop()

            if not node: continue

            if flag==1: res.append(node.val)

            else:

                stack.append((1,node))

                stack.append((0,node.right))

                stack.append((0,node.left))

        return res

递归解法：

class Solution:

    def postorderTraversal(self, root: TreeNode) -> List[int]:

        res= []

        def postorder(root):

            nonlocal res

            if not root:

                return

            postorder(root.left)

            postorder(root.right)

            res.append(root.val)

        postorder(root)

        return res

---

层序遍历：广度优先算法，用队列

Leetcode102. 二叉树的层序遍历

class Solution:

    def levelOrder(self, root: TreeNode) -> List[List[int]]:

        if not root: return []

        res, q= [], [root]

        while q:

            level=[]

            n= len(q)

            for i in range(n):

                node=q.pop(0)

                level.append(node.val)

                if node.left:

                    q.append(node.left)

                if node.right:

                    q.append(node.right)

            res.append(level)

        return res

---

伪码：

递归解法

迭代解法