[Leetcode]Minimum Depth of Binary Tree

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

判断树的最小深度~树的最小深度指的是从根节点到最近的叶节点的距离，所以如果某一个节点只有左子树或右子树，我们不能取它左右子树中小的作为深度，因为其并非叶节点，要注意处理这种情况~下面代码是递归的解法~

class Solution:    # @param root, a tree node    # @return an integer    def minDepth(self, root):        if root is None: return 0        if root.left is None: return self.minDepth(root.right) + 1        if root.right is None: return self.minDepth(root.left) + 1        return min(self.minDepth(root.left), self.minDepth(root.right)) + 1

知道递归解法还不够，最好也能写出非递归解法~代码如下~(注意：python中list.pop(0)时间复杂度为O(n)，所以下面curr = queue.pop(0)这句不是很高效)

class Solution:    # @param root, a tree node    # @return an integer    def minDepth(self, root):        if root is None: return 0        queue = [root, None]; depth = 1        while queue:            curr = queue.pop(0)            if curr == None:                # note: do not increase depth if queue                #   is empty;                 if queue:                    queue.append(None)                    depth += 1            else:                if curr.left is None and curr.right is None:                    return depth                if curr.left:                    queue.append(curr.left)                if curr.right:                    queue.append(curr.right)