Given the root of a binary tree, then value v and depth d, you need to add a row of nodes with value v at the given depth d. The root node is at depth 1. 
The adding rule is: given a positive integer depth d, for each NOT null tree nodes N in depth d-1, create two tree nodes with value v as N's left subtree root and right subtree root. And N's original left subtree should be the left subtree of the new left subtree root, its original right subtree should be the right subtree of the new right subtree root. If depth d is 1 that means there is no depth d-1 at all, then create a tree node with value v as the new root of the whole original tree, and the original tree is the new root's left subtree.
Example 1:
Input: 
A binary tree as following:
       4
     /   \
    2     6
   / \   / 
  3   1 5   
v = 1
d = 2
Output: 
       4
      / \
     1   1
    /     \
   2       6
  / \     / 
 3   1   5   
Example 2:
Input: 
A binary tree as following:
      4
     /   
    2    
   / \   
  3   1    
v = 1
d = 3
Output: 
      4
     /   
    2
   / \    
  1   1
 /     \  
3       1
Note:
The given d is in range [1, maximum depth of the given tree + 1].
The given binary tree has at least one tree node.
class Solution:
     def addOneRow(self, root, v, d):
         """
         :type root: TreeNode
         :type v: int
         :type d: int
         :rtype: TreeNode
         """
         # if d == 1: # 广度搜索 64ms
         #     newroot = TreeNode(v)
         #     newroot.left = root
         #     return newroot
         # lastlevel = []
         # cur = 1
         # curlevel = [root]
         # nextlevel = []
         # while cur < d:
         #     for node in curlevel:
         #         if node:
         #             nextlevel.append(node.left)
         #             nextlevel.append(node.right)
         #     lastlevel = curlevel
         #     curlevel = nextlevel
         #     nextlevel = []
         #     cur += 1
         # count = 0
         # for node in lastlevel:
         #     if node:
         #         templeft, tempright = TreeNode(v), TreeNode(v)
         #         templeft.left = curlevel[count]
         #         tempright.right = curlevel[count+1]
         #         count += 2
         #         node.left = templeft
         #         node.right = tempright
         # return root
         
         if d == 1:
             newroot = TreeNode(v)
             newroot.left = root
             return newroot
         queue = [(root, 1)]
         for node, level in queue:
             if node.left:
                 queue.append((node.left, level+1))
             if node.right:
                 queue.append((node.right, level+1))
             if level == d-1:
                 newleft = TreeNode(v)
                 newleft.left = node.left
                 newright = TreeNode(v)
                 newright.right = node.right
                 node.left = newleft
                 node.right = newright
         return root
class Solution:
     def addOneRow(self, root, v, d):
         """
         :type root: TreeNode
         :type v: int
         :type d: int
         :rtype: TreeNode
         """
         # if d == 1: # 广度搜索 64ms
         #     newroot = TreeNode(v)
         #     newroot.left = root
         #     return newroot
         # lastlevel = []
         # cur = 1
         # curlevel = [root]
         # nextlevel = []
         # while cur < d:
         #     for node in curlevel:
         #         if node:
         #             nextlevel.append(node.left)
         #             nextlevel.append(node.right)
         #     lastlevel = curlevel
         #     curlevel = nextlevel
         #     nextlevel = []
         #     cur += 1
         # count = 0
         # for node in lastlevel:
         #     if node:
         #         templeft, tempright = TreeNode(v), TreeNode(v)
         #         templeft.left = curlevel[count]
         #         tempright.right = curlevel[count+1]
         #         count += 2
         #         node.left = templeft
         #         node.right = tempright
         # return root
         
         if d == 1:
             newroot = TreeNode(v)
             newroot.left = root
             return newroot
         queue = [(root, 1)]
         for node, level in queue:
             if node.left:
                 queue.append((node.left, level+1))
             if node.right:
                 queue.append((node.right, level+1))
             if level == d-1:
                 newleft = TreeNode(v)
                 newleft.left = node.left
                 newright = TreeNode(v)
                 newright.right = node.right
                 node.left = newleft
                 node.right = newright
         return root