LeetCode: 145. Binary Tree Postorder Traversal

题目

原题地址：https://leetcode.com/problems/binary-tree-postorder-traversal/

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

Example 1:

image1

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

Example 2:

Input: root = []
Output: []

Example 3:

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

Example 4:

image4

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

Example 5:

image5

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

Constraints:

  • The number of 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?

解法

递归法实现后序遍历

这个方法的 Python 代码类似下面这样:

# 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):
        nodes = []
        self._postorder(nodes, root)
        return nodes

    def _postorder(self, nodes, root):
        if root is None:
            return

        self._postorder(nodes, root.left)
        self._postorder(nodes, root.right)
        nodes.append(root.val)

stack 法实现后序遍历

这个方法的 Python 代码类似下面这样:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def postorderTraversal(self, root):
        if root is None:
            return []

        nodes = []
        stack1 = []
        stack2 = []

        stack1.append(root)
        while stack1:
            curr = stack1.pop()
            stack2.append(curr)
            if curr.left is not None:
                stack1.append(curr.left)
            if curr.right is not None:
                stack1.append(curr.right)

        while stack2:
            node = stack2.pop()
            nodes.append(node.val)

        return nodes

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