145.Binary_Tree_Postorder_Traversal.md

145. Binary Tree Postorder Traversal

简单

https://leetcode-cn.com/problems/binary-tree-postorder-traversal/

Given the root of a binary tree, return the postorder traversal of its nodes' values.

Example 1:


Input: root = [1,null,2,3]
Output: [3,2,1]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Constraints:

The number of the nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100

Follow up: Recursive solution is trivial, could you do it iteratively?

相关企业

• Facebook|3
• 字节跳动|2

相关标签

• Stack
• Tree
• Depth-First Search
• Binary Tree

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# 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: Optional[TreeNode]) -> List[int]:
        if not root:
            return []

        result = []
        result.extend(self.postorderTraversal(root.left))
        result.extend(self.postorderTraversal(root.right))
        result.append(root.val)
        return result