代码随想录算法训练营第十四天| 二叉树的递归遍历，迭代遍历

递归遍历

144.二叉树的前序遍历

题目链接：https://leetcode.cn/problems/binary-tree-preorder-traversal/

class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []

        def tran(root: TreeNode):
            if root is None:
                return None
            result.append(root.val)
            tran(root.left)
            tran(root.right)

        tran(root)
        return result

145.二叉树的后序遍历

题目链接： https://leetcode.cn/problems/binary-tree-postorder-traversal/

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

        result = []

        def tran(root: TreeNode):
            if root is None:
                return None
            tran(root.left)
            tran(root.right)
            result.append(root.val)
        
        tran(root)
        return result

94.二叉树的中序遍历

题目链接： https://leetcode.cn/problems/binary-tree-inorder-traversal/

class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []

        def tran(root: TreeNode):
            if root is None:
                return None
            tran(root.left)
            result.append(root.val)
            tran(root.right)

        tran(root)
        return result

迭代遍历

题目一样 直接提供代码

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

            if node.right:
                stack.append(node.right)
            if node.left:
                stack.append(node.left)
        return result

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

        if not root:
            return []
        stack = [root]
        result = []

        while stack:
            node = stack.pop()

            result.append(node.val)

            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)
        return result[::-1]

class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        
        stack = []
        result = []
        cur = root
        while cur or stack:
            if cur:
                stack.append(cur)
                cur = cur.left
            else:
                cur = stack.pop()
                result.append(cur.val)
                cur = cur.right
        return result