Maximum Depth of Binary Tree
最后更新于：2022-04-02 01:11:17
# 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 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 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 q = new LinkedList();
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),
';