Question

Source

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

Problem

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)

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
  vector<int> inorderTraversal(TreeNode* root) {
    vector<int> result;
    helper(root, result);
    return result;
  }

private:
  void helper(TreeNode *root, vector<int> &ret) {
    if (root != NULL) {
      helper(root->left, ret);
      ret.push_back(root->val);
      helper(root->right, ret);
    }
  }
};

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
  public List<Integer> inorderTraversal(TreeNode root) {
    List<Integer> result = new ArrayList<Integer>();
    helper(root, result);
    return result;
  }

  private void helper(TreeNode root, List<Integer> ret) {
    if (root != null) {
      helper(root.left, ret);
      ret.add(root.val);
      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)O(n)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

C++

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *   int val;
 *   TreeNode *left, *right;
 *   TreeNode(int val) {
 *     this->val = val;
 *     this->left = this->right = NULL;
 *   }
 * }
 */
class Solution {
  /**
   * @param root: The root of binary tree.
   * @return: Inorder in vector which contains node values.
   */
public:
  vector<int> inorderTraversal(TreeNode *root) {
    vector<int> result;
    stack<TreeNode *> s;

    while (!s.empty() || NULL != root) {
      if (root != NULL) {
        s.push(root);
        root = root->left;
      } else {
        root = s.top();
        s.pop();
        result.push_back(root->val);
        root = root->right;
      }
    }

    return result;
  }
};

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
  public List<Integer> inorderTraversal(TreeNode root) {
    List<Integer> result = new ArrayList<Integer>();
    if (root == null) return result;

    Deque<TreeNode> stack = new ArrayDeque<TreeNode>();
    while (root != null || (!stack.isEmpty())) {
      if (root != null) {
        stack.push(root);
        root = root.left;
      } else {
        root = stack.pop();
        result.add(root.val);
        root = root.right;
      }
    }

    return result;
  }
}

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)O(n)O(n), 时间复杂度 O(n)O(n)O(n).

Reference