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 . 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。
- 首先需要一直對左子樹迭代並將非空節點壓入 stack
- 節點指針爲空後不再壓入 stack
- 當前節點爲空時進行出 stack 操作,並訪問 stack 頂節點
- 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 , 時間複雜度 .