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?

題解1 - 遞迴版

中序遍歷的訪問順序爲『先左再根後右』，遞迴版最好理解，遞迴調用時注意返回值和遞迴左右子樹的順序即可。

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);
        }
    }
}

源碼分析

Python 這種動態語言在寫遞迴時返回結果好處理點，無需聲明類型。通用的方法爲在遞迴函數入口參數中傳入返回結果，
也可使用分治的方法替代輔助函數。

複雜度分析

樹中每個節點都需要被訪問常數次，時間複雜度近似爲 O(n). 未使用額外輔助空間。

題解2 - 迭代版

使用輔助 stack 改寫遞迴程序，中序遍歷沒有前序遍歷好寫，其中之一就在於出入 stack 的順序和限制規則。我們採用「左根右」的訪問順序可知主要由如下四步構成。

1. 首先需要一直對左子樹迭代並將非空節點壓入 stack
2. 節點指針爲空後不再壓入 stack
3. 當前節點爲空時進行出 stack 操作，並訪問 stack 頂節點
4. 將當前指針p用其右子節點替代

步驟2,3,4對應「左根右」的遍歷結構，只是此時的步驟2取的左值爲空。

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;
    }
}

源碼分析

使用 stack 的思想模擬遞迴，注意迭代的演進和邊界條件即可。

複雜度分析

最壞情況下 stack 保存所有節點，空間複雜度 O(n), 時間複雜度 O(n).

Reference