# Binary Tree Inorder Traversal

## Question

### 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

### 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。

## 題解2 - 迭代版

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

### 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;
}
}


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