README.md

Idea

This problem is quite similar to Maximum Depth of Binary Tree, we also attack this problem using two approach BFS and DFS.

BFS

Since BFS runs layer-wise, so once we experience a leave node not node.left and not node.right == 1, we just return the level of this node.

DFS

Compared to Maximum Depth of Binary Tree, we can come up with a naive recursive algorithm:

if not root:
		return 0
return 1 + min(self.minDepth(root.left), self.minDepth(root.right))

Note: there will be probelm when node have only 1 child, the algorithm will return 0 which is not correct.

A modified version:

if not root:
		return 0
if not root.left or not root.right:
		return 1 + max(self.minDepth(root.left), self.minDepth(root.right))
return 1 + min(self.minDepth(root.left), self.minDepth(root.right))