Binary Tree Inorder Traversal
最后更新于：2022-04-02 01:11:08
# Binary Tree Inorder Traversal
### Source
- leetcode: [Binary Tree Inorder Traversal | LeetCode OJ](https://leetcode.com/problems/binary-tree-inorder-traversal/)
- lintcode: [(67) Binary Tree Inorder Traversal](http://www.lintcode.com/en/problem/binary-tree-inorder-traversal/)
~~~
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 inorderTraversal(TreeNode* root) {
vector result;
helper(root, result);
return result;
}
private:
void helper(TreeNode *root, vector &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 inorderTraversal(TreeNode root) {
List result = new ArrayList();
helper(root, result);
return result;
}
private void helper(TreeNode root, List 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
1. 节点指针为空后不再入栈
1. 当前节点为空时进行出栈操作，并访问栈顶节点
1. 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 inorderTraversal(TreeNode *root) {
vector result;
stack 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 inorderTraversal(TreeNode root) {
List result = new ArrayList();
Stack s = new Stack();
while (root != null || !s.empty()) {
if (root != null) {
s.push(root);
root = root.left;
} else {
root = s.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
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