题解 | #二叉树的后序遍历#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param root TreeNode类 
# @return int整型一维数组
#
class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:  #迭代法1
        if not root:
            return []
        stack = [root]
        result = []
        while stack:
            node = stack.pop()
            # 中结点先处理
            result.append(node.val)
            # 左孩子先入栈
            if node.left:
                stack.append(node.left)
            # 右孩子后入栈
            if node.right:
                stack.append(node.right)
        # 将最终的数组翻转
        return result[::-1]

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:  #迭代法2
        result = []
        st = []
        if root:
            st.append(root)
        while st:
            node = st.pop()
            if node != None:
                st.append(node) #中
                st.append(None)
                
                if node.right: #右
                    st.append(node.right)
                if node.left: #左
                    st.append(node.left)
            else:
                node = st.pop()
                result.append(node.val)
        return result

class Solution:
    def postorderTraversal(self , root: TreeNode) -> List[int]:  #递归法
        # write code here
        if root is None: return None
        result = []
        def traversal(root: TreeNode):
            if root is None:
                return None
            if root.left: traversal(root.left)
            if root.right: traversal(root.right)
            result.append(root.val)
        traversal(root)
        return result