问题类型：Easy, BFS
问题描述：
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.
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 minimum depth = 2.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def minDepth(self, root):
"""
:type root: TreeNode
:rtype: int
"""
elements = [[root, 1]]
record = {}
depth = 1
while elements != []:
[cur, depth] = elements.pop(0)
if cur != None:
if cur.left == None and cur.right == None:
return depth
elements += [[cur.left, depth + 1], [cur.right, depth + 1]]
return 0
解决思路：主要逻辑还是通过使用队列进行树的层序遍历，注意，层序即深度。那按照层序遍历的逻辑，从根部开始，节点非空，就把节点的左右子孙加到队列，然后加一个判断，看节点是否为叶子节点。最主要的是，遍历过程中记录一下深度/层序。这就是这类题目的组合逻辑，本身不复杂。
END.
