# Maximum Depth of Binary Tree
### Source
- lintcode: [(97) Maximum Depth of Binary Tree](http://www.lintcode.com/en/problem/maximum-depth-of-binary-tree/)
### Problem
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the rootnode down to the farthest leaf node.
#### Example
Given a binary tree as follow:
~~~
1
/ \
2 3
/ \
4 5
~~~
The maximum depth is `3`.
### 题解 - 递归
树遍历的题最方便的写法自然是递归，不过递归调用的层数过多可能会导致栈空间溢出，因此需要适当考虑递归调用的层数。我们首先来看看使用递归如何解这道题，要求二叉树的最大深度，直观上来讲使用深度优先搜索判断左右子树的深度孰大孰小即可，从根节点往下一层树的深度即自增1，遇到`NULL`时即返回0。
由于对每个节点都会使用一次`maxDepth`，故时间复杂度为 O(n)O(n)O(n), 树的深度最大为 nnn, 最小为 log2n\log_2 nlog2n, 故空间复杂度介于 O(logn)O(\log n)O(logn) 和 O(n)O(n)O(n) 之间。
### 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: An integer
*/
int maxDepth(TreeNode *root) {
if (NULL == root) {
return 0;
}
int left_depth = maxDepth(root->left);
int right_depth = maxDepth(root->right);
return max(left_depth, right_depth) + 1;
}
};
~~~
### Java
~~~
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: An integer.
*/
public int maxDepth(TreeNode root) {
// write your code here
if (root == null) {
return 0;
}
return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
}
}
~~~
### 题解 - 迭代(显式栈)
使用递归可能会导致栈空间溢出，这里使用显式栈空间(使用堆内存)来代替之前的隐式栈空间。从上节递归版的代码(先处理左子树，后处理右子树，最后返回其中的较大值)来看，是可以使用类似后序遍历的迭代思想去实现的。
首先使用后序遍历的模板，在每次迭代循环结束处比较栈当前的大小和当前最大值`max_depth`进行比较。
### 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: An integer
*/
int maxDepth(TreeNode *root) {
if (NULL == root) {
return 0;
}
TreeNode *curr = NULL, *prev = NULL;
stack<TreeNode *> s;
s.push(root);
int max_depth = 0;
while(!s.empty()) {
curr = s.top();
if (!prev || prev->left == curr || prev->right == curr) {
if (curr->left) {
s.push(curr->left);
} else if (curr->right){
s.push(curr->right);
}
} else if (curr->left == prev) {
if (curr->right) {
s.push(curr->right);
}
} else {
s.pop();
}
prev = curr;
if (s.size() > max_depth) {
max_depth = s.size();
}
}
return max_depth;
}
};
~~~
### 题解3 - 迭代(队列)
在使用了递归/后序遍历求解树最大深度之后，我们还可以直接从问题出发进行分析，树的最大深度即为广度优先搜索中的层数，故可以直接使用广度优先搜索求出最大深度。
### 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: An integer
*/
int maxDepth(TreeNode *root) {
if (NULL == root) {
return 0;
}
queue<TreeNode *> q;
q.push(root);
int max_depth = 0;
while(!q.empty()) {
int size = q.size();
for (int i = 0; i != size; ++i) {
TreeNode *node = q.front();
q.pop();
if (node->left) {
q.push(node->left);
}
if (node->right) {
q.push(node->right);
}
}
++max_depth;
}
return max_depth;
}
};
~~~
### Java
~~~
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: An integer.
*/
public int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}
int depth = 0;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
while (!q.isEmpty()) {
depth++;
int qLen = q.size();
for (int i = 0; i < qLen; i++) {
TreeNode node = q.poll();
if (node.left != null) q.offer(node.left);
if (node.right != null) q.offer(node.right);
}
}
return depth;
}
}
~~~
### 源码分析
广度优先中队列的使用中，`qLen` 需要在for 循环遍历之前获得，因为它是一个变量。
### 复杂度分析
最坏情况下空间复杂度为 O(n)O(n)O(n), 遍历每一个节点，时间复杂度为 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