# Maximum Depth of Binary Tree

`Tree`, `BFS`, `DFS`

Easy

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

**Note:** A leaf is a node with no children.

**Example:**

Given binary tree`[3,9,20,null,null,15,7]`,

```
    3
   / \
  9  20
    /  \
   15   7
```

return its depth = 3.

## Solution

### **\*Recursive**

```java
    public int maxDepth(TreeNode root) {
        if(root==null){
            return 0;
        }
        return 1+Math.max(maxDepth(root.left),maxDepth(root.right));
    }
```

### DFS

```java
public int maxDepth(TreeNode root) {
    if(root == null) {
        return 0;
    }

    Stack<TreeNode> stack = new Stack<>();
    Stack<Integer> value = new Stack<>();
    stack.push(root);
    value.push(1);
    int max = 0;
    while(!stack.isEmpty()) {
        TreeNode node = stack.pop();
        int temp = value.pop();
        max = Math.max(temp, max);
        if(node.left != null) {
            stack.push(node.left);
            value.push(temp+1);
        }
        if(node.right != null) {
            stack.push(node.right);
            value.push(temp+1);
        }
    }
    return max;
}
```

### \*BFS

```java
public int maxDepth(TreeNode root) {
    if(root == null) {
        return 0;
    }
    Queue<TreeNode> queue = new LinkedList<>();
    queue.offer(root);
    int count = 0;
    while(!queue.isEmpty()) {
        int size = queue.size();
        while(size-- > 0) {
            TreeNode node = queue.poll();
            if(node.left != null) {
                queue.offer(node.left);
            }
            if(node.right != null) {
                queue.offer(node.right);
            }
        }
        count++;
    }
    return count;
}
```


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