【leetcode】623. Add One Row to Tree

题目如下：

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 subtreeshould 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:

1. The given d is in range [1, maximum depth of the given tree + 1].
2. The given binary tree has at least one tree node.

解题思路：本题不难，我的方法是用BFS遍历整个二叉树，遍历过程中记录节点的level；如果level+1小于d，把node的子节点加入queue中；如果level+1等于d，表示要在这个节点下面插入两个左右子节点，只需创建出两个子节点，让其中一个的left指向node.left，另一个的right指向node.right，之后再让node.left和node.right分别指向这两个新创建的节点即可。

代码如下：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def addOneRow(self, root, v, d):
        """
        :type root: TreeNode
        :type v: int
        :type d: int
        :rtype: TreeNode
        """
        if d == 1:
            node =  TreeNode(v)
            node.left = root
            return node
        queue = [(root,1)]
        while len(queue) > 0:
            node,level = queue.pop(0)
            if level >= d:
                break
            elif level + 1 == d:
                l_node = TreeNode(v)
                r_node = TreeNode(v)
                l_node.left = node.left
                r_node.right = node.right
                node.left = l_node
                node.right = r_node
            else:
                if node.left != None:
                    queue.append((node.left,level+1))
                if node.right != None:
                    queue.append((node.right, level + 1))
        return root