**1.if we use simple divide and conquer technique(without using extra space)**

we first need to count no. of leaf in left subtree,then no. of leaf in right subtree+compare them to find minimum.

**T(n)=2T(n/2)+1----->O(n)**

but **we need to compute this for every node.**

so, cost at 0 th level=O(n)

cost at 1st level=n/2+n/2=O(n)

cost at 2nd level=n/4+n/4+n/4+n/4=O(n)\

we have total of **logn** levels (because tree is balanced) and cost at every level is O(n).

Hence total cost is=n*logn=O(nlogn)

**2.if we use extra memory to store no of leaf at left subtree and at right subtree for every node**

we don't need to go to last level for every node.

1.compute for every leaf node.

2.then use those values to compute at all nodes at (h-1) level till root.

so,** time complexity** of this method is equal to** O(n)**,since we are calculating value for each node and doing only one comparison.

**space complexity** is also** O(n)**,since we are using extra memory for every node.

since question is asking about** worst case time complexity**,so we have **O(nlogn)** .