Problem
Given the root of a binary tree, return the postorder traversal of its nodes' values.
Example 1:
img
Input: root = [1,null,2,3]
Output: [3,2,1]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Example 4:
img
Input: root = [1,2]
Output: [2,1]
Example 5:
img
Input: root = [1,null,2]
Output: [2,1]
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?
问题
给定一个二叉树，返回它的 后序 遍历。
示例:
输入: [1,null,2,3]
1
\
2
/
3
输出: [3,2,1]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？
思路
递归
左右根
先遍历左右子树，再加入 root 节点的值。
Python3 代码
# 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) -> List[int]:
# 递归
res = []
def dfs(root):
if not root:
return []
dfs(root.left)
dfs(root.right)
res.append(root.val)
dfs(root)
return res
迭代
使用栈来模拟
Python3 代码
# 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) -> List[int]:
# 迭代
res = []
if not root:
return res
stack = []
node = root
while stack or node:
while node:
# 根节点入栈
stack.append(node)
# 左子树存在
if node.left:
node = node.left
# 左子树不存在，转右子树
else:
node = node.right
# 取出栈顶元素
node = stack.pop()
res.append(node.val)
# 栈不为空且当前节点是栈顶元素的左节点
# stack[-1]就是取出的栈顶元素：node
if stack and node == stack[-1].left:
node = stack[-1].right
# 没有左子树或右子树，退栈
else:
node = None
return res
