# Binary Tree Preorder Traversal
### Source
- leetcode: [Binary Tree Preorder Traversal | LeetCode OJ](https://leetcode.com/problems/binary-tree-preorder-traversal/)
- lintcode: [(66) Binary Tree Preorder Traversal](http://www.lintcode.com/en/problem/binary-tree-preorder-traversal/)
~~~
Given a binary tree, return the preorder traversal of its nodes' values.
Note
Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,2,3].
Example
Challenge
Can you do it without recursion?
~~~
### 题解1 - 递归
**面试时不推荐递归这种做法。**
递归版很好理解，首先判断当前节点(根节点)是否为`null`，是则返回空vector，否则先返回当前节点的值，然后对当前节点的左节点递归，最后对当前节点的右节点递归。递归时对返回结果的处理方式不同可进一步细分为遍历和分治两种方法。
### Python - Divide and Conquer
~~~
"""
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: Preorder in ArrayList which contains node values.
"""
def preorderTraversal(self, root):
if root == None:
return []
return [root.val] + self.preorderTraversal(root.left) \
+ self.preorderTraversal(root.right)
~~~
### C++ - Divide and Conquer
~~~
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param root: The root of binary tree.
* @return: Preorder in vector which contains node values.
*/
vector<int> preorderTraversal(TreeNode *root) {
vector<int> result;
if (root != NULL) {
// Divide (分)
vector<int> left = preorderTraversal(root->left);
vector<int> right = preorderTraversal(root->right);
// Merge
result.push_back(root->val);
result.insert(result.end(), left.begin(), left.end());
result.insert(result.end(), right.begin(), right.end());
}
return result;
}
};
~~~
### C++ - Traversal
~~~
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param root: The root of binary tree.
* @return: Preorder in vector which contains node values.
*/
vector<int> preorderTraversal(TreeNode *root) {
vector<int> result;
traverse(root, result);
return result;
}
private:
void traverse(TreeNode *root, vector<int> &ret) {
if (root != NULL) {
ret.push_back(root->val);
traverse(root->left, ret);
traverse(root->right, ret);
}
}
};
~~~
### Java - Divide and Conquer
~~~
/**
* 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> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
if (root != null) {
// Divide
List<Integer> left = preorderTraversal(root.left);
List<Integer> right = preorderTraversal(root.right);
// Merge
result.add(root.val);
result.addAll(left);
result.addAll(right);
}
return result;
}
}
~~~
### 源码分析
使用遍历的方法保存递归返回结果需要使用辅助递归函数`traverse`，将结果作为参数传入递归函数中，传值时注意应使用`vector`的引用。分治方法首先分开计算各结果，最后合并到最终结果中。C++ 中由于是使用vector, 将新的vector插入另一vector不能再使用push_back, 而应该使用insert。Java 中使用`addAll`方法.
### 复杂度分析
遍历树中节点，时间复杂度 O(n)O(n)O(n), 未使用额外空间。
### 题解2 - 迭代
迭代时需要利用栈来保存遍历到的节点，纸上画图分析后发现应首先进行出栈抛出当前节点，保存当前节点的值，随后将右、左节点分别入栈(注意入栈顺序，先右后左)，迭代到其为叶子节点(NULL)为止。
### 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 preorderTraversal(self, root):
if root is None:
return []
result = []
s = []
s.append(root)
while s:
root = s.pop()
result.append(root.val)
if root.right is not None:
s.append(root.right)
if root.left is not None:
s.append(root.left)
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 {
public:
/**
* @param root: The root of binary tree.
* @return: Preorder in vector which contains node values.
*/
vector<int> preorderTraversal(TreeNode *root) {
vector<int> result;
if (root == NULL) return result;
stack<TreeNode *> s;
s.push(root);
while (!s.empty()) {
TreeNode *node = s.top();
s.pop();
result.push_back(node->val);
if (node->right != NULL) {
s.push(node->right);
}
if (node->left != NULL) {
s.push(node->left);
}
}
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> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
if (root == null) return result;
Stack<TreeNode> s = new Stack<TreeNode>();
s.push(root);
while (!s.empty()) {
TreeNode node = s.pop();
result.add(node.val);
if (node.right != null) s.push(node.right);
if (node.left != null) s.push(node.left);
}
return result;
}
}
~~~
### 源码分析
1. 对root进行异常处理
1. 将root压入栈
1. 循环终止条件为栈s为空，所有元素均已处理完
1. 访问当前栈顶元素(首先取出栈顶元素，随后pop掉栈顶元素)并存入最终结果
1. 将右、左节点分别压入栈内，以便取元素时为先左后右。
1. 返回最终结果
其中步骤4,5,6为迭代的核心，对应前序遍历「根左右」。
所以说到底，**使用迭代，只不过是另外一种形式的递归。**使用递归的思想去理解遍历问题会容易理解许多。
### 复杂度分析
使用辅助栈，最坏情况下栈空间与节点数相等，空间复杂度近似为 O(n)O(n)O(n), 对每个节点遍历一次，时间复杂度近似为 O(n)O(n)O(n).
- Preface
- Part I - Basics
- Basics Data Structure
- String
- Linked List
- Binary Tree
- Huffman Compression
- Queue
- Heap
- Stack
- Set
- Map
- Graph
- Basics Sorting
- Bubble Sort
- Selection Sort
- Insertion Sort
- Merge Sort
- Quick Sort
- Heap Sort
- Bucket Sort
- Counting Sort
- Radix Sort
- Basics Algorithm
- Divide and Conquer
- Binary Search
- Math
- Greatest Common Divisor
- Prime
- Knapsack
- Probability
- Shuffle
- Basics Misc
- Bit Manipulation
- Part II - Coding
- String
- strStr
- Two Strings Are Anagrams
- Compare Strings
- Anagrams
- Longest Common Substring
- Rotate String
- Reverse Words in a String
- Valid Palindrome
- Longest Palindromic Substring
- Space Replacement
- Wildcard Matching
- Length of Last Word
- Count and Say
- Integer Array
- Remove Element
- Zero Sum Subarray
- Subarray Sum K
- Subarray Sum Closest
- Recover Rotated Sorted Array
- Product of Array Exclude Itself
- Partition Array
- First Missing Positive
- 2 Sum
- 3 Sum
- 3 Sum Closest
- Remove Duplicates from Sorted Array
- Remove Duplicates from Sorted Array II
- Merge Sorted Array
- Merge Sorted Array II
- Median
- Partition Array by Odd and Even
- Kth Largest Element
- Binary Search
- Binary Search
- Search Insert Position
- Search for a Range
- First Bad Version
- Search a 2D Matrix
- Search a 2D Matrix II
- Find Peak Element
- Search in Rotated Sorted Array
- Search in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array
- Find Minimum in Rotated Sorted Array II
- Median of two Sorted Arrays
- Sqrt x
- Wood Cut
- Math and Bit Manipulation
- Single Number
- Single Number II
- Single Number III
- O1 Check Power of 2
- Convert Integer A to Integer B
- Factorial Trailing Zeroes
- Unique Binary Search Trees
- Update Bits
- Fast Power
- Hash Function
- Count 1 in Binary
- Fibonacci
- A plus B Problem
- Print Numbers by Recursion
- Majority Number
- Majority Number II
- Majority Number III
- Digit Counts
- Ugly Number
- Plus One
- Linked List
- Remove Duplicates from Sorted List
- Remove Duplicates from Sorted List II
- Remove Duplicates from Unsorted List
- Partition List
- Two Lists Sum
- Two Lists Sum Advanced
- Remove Nth Node From End of List
- Linked List Cycle
- Linked List Cycle II
- Reverse Linked List
- Reverse Linked List II
- Merge Two Sorted Lists
- Merge k Sorted Lists
- Reorder List
- Copy List with Random Pointer
- Sort List
- Insertion Sort List
- Check if a singly linked list is palindrome
- Delete Node in the Middle of Singly Linked List
- Rotate List
- Swap Nodes in Pairs
- Remove Linked List Elements
- Binary Tree
- Binary Tree Preorder Traversal
- Binary Tree Inorder Traversal
- Binary Tree Postorder Traversal
- Binary Tree Level Order Traversal
- Binary Tree Level Order Traversal II
- Maximum Depth of Binary Tree
- Balanced Binary Tree
- Binary Tree Maximum Path Sum
- Lowest Common Ancestor
- Invert Binary Tree
- Diameter of a Binary Tree
- Construct Binary Tree from Preorder and Inorder Traversal
- Construct Binary Tree from Inorder and Postorder Traversal
- Subtree
- Binary Tree Zigzag Level Order Traversal
- Binary Tree Serialization
- Binary Search Tree
- Insert Node in a Binary Search Tree
- Validate Binary Search Tree
- Search Range in Binary Search Tree
- Convert Sorted Array to Binary Search Tree
- Convert Sorted List to Binary Search Tree
- Binary Search Tree Iterator
- Exhaustive Search
- Subsets
- Unique Subsets
- Permutations
- Unique Permutations
- Next Permutation
- Previous Permuation
- Unique Binary Search Trees II
- Permutation Index
- Permutation Index II
- Permutation Sequence
- Palindrome Partitioning
- Combinations
- Combination Sum
- Combination Sum II
- Minimum Depth of Binary Tree
- Word Search
- Dynamic Programming
- Triangle
- Backpack
- Backpack II
- Minimum Path Sum
- Unique Paths
- Unique Paths II
- Climbing Stairs
- Jump Game
- Word Break
- Longest Increasing Subsequence
- Palindrome Partitioning II
- Longest Common Subsequence
- Edit Distance
- Jump Game II
- Best Time to Buy and Sell Stock
- Best Time to Buy and Sell Stock II
- Best Time to Buy and Sell Stock III
- Best Time to Buy and Sell Stock IV
- Distinct Subsequences
- Interleaving String
- Maximum Subarray
- Maximum Subarray II
- Longest Increasing Continuous subsequence
- Longest Increasing Continuous subsequence II
- Graph
- Find the Connected Component in the Undirected Graph
- Route Between Two Nodes in Graph
- Topological Sorting
- Word Ladder
- Bipartial Graph Part I
- Data Structure
- Implement Queue by Two Stacks
- Min Stack
- Sliding Window Maximum
- Longest Words
- Heapify
- Problem Misc
- Nuts and Bolts Problem
- String to Integer
- Insert Interval
- Merge Intervals
- Minimum Subarray
- Matrix Zigzag Traversal
- Valid Sudoku
- Add Binary
- Reverse Integer
- Gray Code
- Find the Missing Number
- Minimum Window Substring
- Continuous Subarray Sum
- Continuous Subarray Sum II
- Longest Consecutive Sequence
- Part III - Contest
- Google APAC
- APAC 2015 Round B
- Problem A. Password Attacker
- Microsoft
- Microsoft 2015 April
- Problem A. Magic Box
- Problem B. Professor Q's Software
- Problem C. Islands Travel
- Problem D. Recruitment
- Microsoft 2015 April 2
- Problem A. Lucky Substrings
- Problem B. Numeric Keypad
- Problem C. Spring Outing
- Microsoft 2015 September 2
- Problem A. Farthest Point
- Appendix I Interview and Resume
- Interview
- Resume