LeetCode145:二叉树的后序遍历(递归+非递归)

目录

一、题目

二、示例

三、思路

四、代码


一、题目

给定一个二叉树,返回它的 后序遍历

二、示例

示例:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [3,2,1]

三、思路

1、递归。后序遍历:左-->右-->根

2、非递归,的思想,先进后出。

四、代码

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 postorderTraversal(self, root: TreeNode):
        if root is None:
            return []
        return self.postorderTraversal(root.left) + self.postorderTraversal(root.right) + [root.val]

if __name__ == '__main__':
    root = TreeNode(1)
    root.left = None
    root.right = TreeNode(2)
    root.right.left = TreeNode(3)
    s = Solution()
    ans = s.postorderTraversal(root)
    print(ans)

2、

# 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

    def postorderTraversal(self, root: TreeNode):
        res = []
        if root is None:
            return res
        stack = [root]

        while (stack):
            root = stack.pop()
            if root is None:
                continue
            stack.append(root.left)
            stack.append(root.right)
            res.append(root.val)
        res.reverse()
        return res

if __name__ == '__main__':
    root = TreeNode(1)
    root.left = None
    root.right = TreeNode(2)
    root.right.left = TreeNode(3)
    s = Solution()
    ans = s.postorderTraversal(root)
    print(ans)

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转载自blog.csdn.net/weixin_45666660/article/details/108739862