题目如下:
Given the root of a binary tree, then value
v
and depthd
, you need to add a row of nodes with valuev
at the given depthd
. The root node is at depth 1.The adding rule is: given a positive integer depth
d
, for each NOT null tree nodesN
in depthd-1
, create two tree nodes with valuev
asN's
left subtree root and right subtree root. AndN'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 depthd
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 5Example 2:
Input:
A binary tree as following:
4
/
2
/ \
3 1 v = 1 d = 3 Output:
4
/
2
/ \
1 1
/ \
3 1Note:
- The given d is in range [1, maximum depth of the given tree + 1].
- 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