# Totally different solution explained (Beats 96% Javascript submissions)

1. Save the height values of all the branches of the tree in an array.
2. Sort the array.
3. Return the first value element of the array.
``````var minDepth = function(root) {
var res = [];

var find = function(root, count){
if(root.left === null & root.right === null)
res.push(count);

if(root.left !== null)
find(root.left, count + 1);
if(root.right !== null)
find(root.right, count + 1);
};

if(root === null)
return 0;

find(root, 1);

res.sort(function(a, b) {
return a - b;
});

return res[0];
};``````

• I took inspiration from your solution except made a slight performance improvement (don't need to sort at the end).

Just keep a running minimum.

``````var minDepth = function(root) {
var minDepth = Infinity;

if (root === null) {
return 0;
}

function find(node, depth = 1) {
if (node.left === null && node.right === null) {
if (depth < minDepth) {
minDepth = depth;
if (minDepth === 1) {
return;
}
}
}

if (node.left !== null) {
find(node.left, depth+1);
}

if (node.right !== null) {
find(node.right, depth+1);
}
}

find(root);

return minDepth;
};

``````

• Save height of all branches is O(n)
Sorting is O(nlogn)

Your solution in (n log n) which is not optimal solution that can be found.

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