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add-one-row-to-tree.py
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add-one-row-to-tree.py
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# Time: O(n)
# Space: O(h)
# 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:
# 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.
# 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 in (0, 1):
node = TreeNode(v)
if d == 1:
node.left = root
else:
node.right = root
return node
if root and d >= 2:
root.left = self.addOneRow(root.left, v, d-1 if d > 2 else 1)
root.right = self.addOneRow(root.right, v, d-1 if d > 2 else 0)
return root