623. Add One Row to Tree 將一行添加到樹中
阿新 • • 發佈:2018-02-02
temp str top spa order app num example des 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
來自為知筆記(Wiz)
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
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 = 1d = 2Output: 4 / 1 1 / 2 6 / \ / 3 1 5
Example 2:
Input: A binary tree as following: 4 / 2 / \ 3 1 v = 1d = 3Output: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 is 1:
newRoot = TreeNode(v)
newRoot.left = root
return newRoot
queue = [root]
while queue and d > 2:
d -= 1
count = len(queue)
for i in range(count):
node = queue.pop(0)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
for curNode in queue:
temp = curNode.left
curNode.left = TreeNode(v)
curNode.left.left = temp
temp = curNode.right
curNode.right = TreeNode(v)
curNode.right.right = temp
return root
來自為知筆記(Wiz)
623. Add One Row to Tree 將一行添加到樹中