2025/5/10 【二叉树】前中后序递归遍历 LeetCode94/ 145/ 144 【√】

看明白了卡哥的讲解后，做的很快，很简单。代码随想录

144. 二叉树的前序遍历 - 力扣（LeetCode）

写法1

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def Traversal(self, root, result):
        if root == None:
            return
        result.append(root.val)
        self.Traversal(root.left, result)
        self.Traversal(root.right, result)

    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        self.Traversal(root, result)
        return result

写法2：简化结构：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

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

这是把递归函数嵌套在 preorderTraversal 里面，减少类中方法的数量，但功能完全一致。

写法3：非递归版本，用栈

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

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

这个用的是手动维护一个栈来模拟递归过程，更接近底层执行逻辑，也适合面试中考察。

94. 二叉树的中序遍历 - 力扣（LeetCode）

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def Traversal(self, root, result):
        if root == None:
            return
        self.Traversal(root.left, result)
        result.append(root.val)
        self.Traversal(root.right, result)

    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        self.Traversal(root, result)
        return result
        

145. 二叉树的后序遍历 - 力扣（LeetCode）

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = rightb
class Solution:
    def Traversal(self, root, result):
        if root == None:
            return
        self.Traversal(root.left, result)
        self.Traversal(root.right, result)
        result.append(root.val)

    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        self.Traversal(root, result)
        return result