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''' | ||
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: | ||
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: | ||
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. | ||
''' | ||
#Difficulty: Medium | ||
#109 / 109 test cases passed. | ||
#Runtime: 56 ms | ||
#Memory Usage: 16.2 MB | ||
|
||
#Runtime: 56 ms, faster than 64.46% of Python3 online submissions for Add One Row to Tree. | ||
#Memory Usage: 16.2 MB, less than 79.05% of Python3 online submissions for Add One Row to Tree. | ||
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||
# Definition for a binary tree node. | ||
# class TreeNode: | ||
# def __init__(self, val=0, left=None, right=None): | ||
# self.val = val | ||
# self.left = left | ||
# self.right = right | ||
|
||
class Solution: | ||
def addOneRow(self, root: TreeNode, v: int, d: int) -> TreeNode: | ||
depth = 1 | ||
queue = [root] | ||
|
||
if depth == d: | ||
left = root | ||
root = TreeNode(v) | ||
root.left = left | ||
|
||
while queue or depth <= d: | ||
length = len(queue) | ||
depth += 1 | ||
while length: | ||
length -= 1 | ||
node = queue.pop(0) | ||
if node: | ||
queue.append(node.left) | ||
queue.append(node.right) | ||
if depth == d and node: | ||
left = node.left | ||
right = node.right | ||
if left: | ||
node.left = TreeNode(v) | ||
node.left.left = left | ||
if right: | ||
node.right = TreeNode(v) | ||
node.right.right = right | ||
if not node.left: | ||
node.left = TreeNode(v) | ||
if not node.right: | ||
node.right = TreeNode(v) | ||
else: | ||
node = TreeNode(v) | ||
return root |
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