# Leetcode Maximum-Depth-Of-Binary-Tree

#### 2022-07-05 09:25

> ##### Algorithms:
> #algorithm #BFS
> ##### Data structures:
> #DS #binary_tree 
> ##### Difficulty:
> #coding_problem #difficulty-easy
> ##### Additional tags:
> #leetcode
> ##### Revisions:
> N/A

##### Related topics:
##### Links:
- [Link to problem](https://leetcode.com/problems/maximum-depth-of-binary-tree/)
___
### Problem
Given the `root` of a binary tree, return _its maximum depth_.

A binary tree's **maximum depth** is the number of nodes along the longest path from the root node down to the farthest leaf node.

#### Examples

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/11/26/tmp-tree.jpg)

**Input:** root = [3,9,20,null,null,15,7]
**Output:** 3

**Example 2:**

**Input:** root = [1,null,2]
**Output:** 2

#### Constraints

-   The number of nodes in the tree is in the range `[0, 104]`.
-   `-100 <= Node.val <= 100`

### Thoughts

> [!summary]
> This problem can be solved by #BFS or #DFS

BFS way:
Simply log the level in each while iteration.

DFS way: (Popular)
Use recursion:
- Base Case:
	- root == nullptr: return 0;
- maxDepth(root) = max(maxDepth(root->left), maxDepth(root->right))

### Solution

DFS Recursion:
```cpp
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
  int maxDepth(TreeNode *root) {
    // DFS
    if (!root) {
      return 0;
    }

    return max(maxDepth(root->left), maxDepth(root->right)) + 1;
  }
};
```

BFS:
```cpp
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
  int maxDepth(TreeNode *root) {
    // BFS
    int levels = 0;
    queue<TreeNode *> pending;
    TreeNode *ptr = root;
    if (ptr)
      pending.push(ptr);

    while (!pending.empty()) {
      levels++;
      for (int i = 0, size = pending.size(); i < size; i++) {
        ptr = pending.front();
        pending.pop();

        if (ptr->left)
          pending.push(ptr->left);
        if (ptr->right)
          pending.push(ptr->right);
      }
    }

    return levels;
  }
};
```