Binary Tree Inorder Traversal

Question

• leetcode: Binary Tree Inorder Traversal | LeetCode OJ
• lintcode: (67) Binary Tree Inorder Traversal

Problem Statement

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

Example

Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

Challenge

Can you do it without recursion?

题解1 - 递归版

中序遍历的访问顺序为『先左再根后右』，递归版最好理解，递归调用时注意返回值和递归左右子树的顺序即可。

Python

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        this.val = val
        this.left, this.right = None, None
"""


class Solution:
    """
    @param root: The root of binary tree.
    @return: Inorder in ArrayList which contains node values.
    """
    def inorderTraversal(self, root):
        if root is None:
            return []
        else:
            return [root.val] + self.inorderTraversal(root.left) \
                              + self.inorderTraversal(root.right)

Python - with helper

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param {TreeNode} root
    # @return {integer[]}
    def inorderTraversal(self, root):
        result = []
        self.helper(root, result)
        return result

    def helper(self, root, ret):
        if root is not None:
            self.helper(root.left, ret)
            ret.append(root.val)
            self.helper(root.right, ret)

源码分析

Python 这种动态语言在写递归时返回结果好处理点，无需声明类型。通用的方法为在递归函数入口参数中传入返回结果， 也可使用分治的方法替代辅助函数。

复杂度分析

树中每个节点都需要被访问常数次，时间复杂度近似为 $$O(n)$$. 未使用额外辅助空间。

题解2 - 迭代版

使用辅助栈改写递归程序，中序遍历没有前序遍历好写，其中之一就在于入栈出栈的顺序和限制规则。我们采用「左根右」的访问顺序可知主要由如下四步构成。

1. 首先需要一直对左子树迭代并将非空节点入栈
2. 节点指针为空后不再入栈
3. 当前节点为空时进行出栈操作，并访问栈顶节点
4. 将当前指针p用其右子节点替代

步骤2,3,4对应「左根右」的遍历结构，只是此时的步骤2取的左值为空。

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param {TreeNode} root
    # @return {integer[]}
    def inorderTraversal(self, root):
        result = []
        s = []
        while root is not None or s:
            if root is not None:
                s.append(root)
                root = root.left
            else:
                root = s.pop()
                result.append(root.val)
                root = root.right

        return result

• reference

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @return a list of integers
    def iterative_inorder(self, root, list):
        stack = []
        while root or stack:
            if root:
                stack.append(root)
                root = root.left
            else:
                root = stack.pop()
                list.append(root.val)
                root = root.right
        return list

    def recursive_inorder(self, root, list):
        if root:
            self.inorder(root.left, list)
            list.append(root.val)
            self.inorder(root.right, list)

    def inorderTraversal(self, root):
        list = []
        self.iterative_inorder(root, list)
        return list

源码分析

使用栈的思想模拟递归，注意迭代的演进和边界条件即可。

复杂度分析

最坏情况下栈保存所有节点，空间复杂度 $$O(n)$$, 时间复杂度 $$O(n)$$.

Reference