# LeetCode 104. Maximum Depth of Binary Tree

## The Problem

Link to original problem on Leetcode

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.

## My Solution

This solution uses recursion.

const maxDepth = root => {
// Terminate immediately if there is no node provided
// This is handy when called on a non-existant child.
if (!root) {
return 0;
}

// Recursively call this function on the children of the node.
// Add a 1, because if this is a leaf, you'd get Math.max(0, 0).
// The 1 represents this level, which will serve as the starting
// point to count each level back to the root.
return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
};

## Others’ Solutions

Solutions page locked, but here is another JS answer that is roughly equivalent to mine. Could be a one-liner, but I find it easier to read as three.

const maxDepth = root =>
!root ? 0 : 1 + Math.max(maxDepth(root.left), maxDepth(root.right));

Another solution that uses iteration instead of recursion:

const maxDepth = root => {
if (!root) return 0;
// Initialize a queue with the root node.
const queue = [root];
// Initialize the tree's depth at 0.
let depth = 0;

while (queue.length !== 0) {
const len = queue.length;

// Iterates over each item in the queue, adding all child nodes
// to the queue. So in the original example, the queue starts at
//  and becomes [9, 20]. Then the depth is iterated by 1.
// Next would be [9, 20] to [15, 7].
// Loop ended when there's nothing left to add to the queue.
for (let i = 0; i < len; i++) {
let item = queue.shift();
if (item.left) queue.push(item.left);
if (item.right) queue.push(item.right);
}
depth++;
}

// Return the deepest level you could hit.
return depth;
};