Sheng – Page 16 – Code Says

Solution to Populating Next Right Pointers in Each Node II by LeetCode

9 Jul

Question: https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/ Question Name: Populating Next Right Pointers in Each Node II The solution is completely the same as the one for Populating Next Right Pointers in Each Node.

Solution to Populating Next Right Pointers in Each Node by LeetCode

9 Jul

Question: https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/ Question Name: Populating Next Right Pointers in Each Node

# Definition for a binary tree node

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

# self.next = None

class Solution:

# @param root, a tree node

# @return nothing

def connect(self, root):

fatherLayerCurrent = root

sonLayerHead = None

sonLayerCurrent = None

# Travel layer by layer as BFS travesal.

while fatherLayerCurrent != None:

# Travel inside a layer from left to right

while fatherLayerCurrent != None:

# Try to access the left son if there is

if fatherLayerCurrent.left != None:

if sonLayerHead == None:

sonLayerHead = fatherLayerCurrent.left

sonLayerCurrent = sonLayerHead

else:

sonLayerCurrent.next = fatherLayerCurrent.left

sonLayerCurrent = sonLayerCurrent.next

# Try to access the right son if there is

if fatherLayerCurrent.right != None:

if sonLayerHead == None:

sonLayerHead = fatherLayerCurrent.right

sonLayerCurrent = sonLayerHead

else:

sonLayerCurrent.next = fatherLayerCurrent.right

sonLayerCurrent = sonLayerCurrent.next

fatherLayerCurrent = fatherLayerCurrent.next

# Prepare to move down to the next layer.

fatherLayerCurrent = sonLayerHead

sonLayerHead =None

return

Solution to Pascal's Triangle II by LeetCode

7 Jul

Question: https://oj.leetcode.com/problems/pascals-triangle-ii/ Question Name: Pascal’s Triangle II

class Solution:

# @return a list of integers

def getRow(self, rowIndex):

''' Ref: http://en.m.wikipedia.org/wiki/Pascal%27s_triangle

Convert this question into:

how to compute the consecutive binomial coefficients.

'''

if rowIndex == 0: return [1]

if rowIndex == 1: return [1, 1]

result = [1]

# The final result is symmetrical. We could only compute the

# the left part, and then append the reversed array of left

# part to the end.

# COmpute the left part of the symmetrical result

nextDivisor = 1

nextMultiplier = rowIndex

for _ in xrange(rowIndex // 2):

nextVal = (result[-1] * nextMultiplier) / nextDivisor

result.append(nextVal)

nextDivisor += 1

nextMultiplier -= 1

# Append the reversed array of left part

if rowIndex % 2 == 1: result.extend(result[::-1])

else: result.extend(result[-2::-1])

return result

Solution to Pascal's Triangle by LeetCode

7 Jul

Question: https://oj.leetcode.com/problems/pascals-triangle/ Question Name: Pascal’s Triangle

class Solution:

# @return a list of lists of integers

def generate(self, numRows):

if numRows == 0: return []

result = [[1]]

while len(result) < numRows:

temp = [1] # Every row starts with 1

for index in xrange(len(result[-1])-1):

temp.append(result[-1][index] + result[-1][index+1])

temp.append(1) # Every row ends with 1

result.append(temp)

return result

Solution to Distinct Subsequences by LeetCode

7 Jul

Question: https://oj.leetcode.com/problems/distinct-subsequences/ Question Name: Distinct Subsequences This is a dynamic programming challenge. Firstly I solved it with recursion. But that solution did not satisfy the time requirement. These two posts give more details about the right solution: LeetCode: Distinct Subsequences（不同子序列的个数）(in … Read More »

Solution to Flatten Binary Tree to Linked List by LeetCode

6 Jul

Question: https://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/ Question Name: Flatten Binary Tree to Linked List The recursive method is easy to understand.

# Definition for a binary tree node

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

class Solution:

# @param root, a tree node

# @return root, a tree node

def flattenHelper(self, root):

if root == None:

return None

else:

left = root.left

right = root.right

root.left = None # Truncate the left subtree

current = root

# Flatten the left subtree

current.right = self.flattenHelper(left)

while current.right != None: current = current.right

# Flatten the right subtree

current.right = self.flattenHelper(right)

return root

# @param root, a tree node

# @return nothing, do it in place

def flatten(self, root):

self.flattenHelper(root)

return

The iterative solution should be more efficient.

# Definition for a binary tree node

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

class Solution:

# @param root, a tree node

# @return nothing, do it in place

def flatten(self, root):

if root == None: return

stack = [root.right, root.left]

current = root

while len(stack) != 0:

nextNode = stack.pop()

if nextNode == None:

continue

else:

# Truncate the left subtree

current.left = None

current.right = nextNode

current = current.right

stack.append(current.right)

stack.append(current.left)

return

Solution to Path Sum II by LeetCode

6 Jul

Question: https://oj.leetcode.com/problems/path-sum-ii/ Question Name: Path Sum II

# Definition for a binary tree node

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

class Solution:

# @param root, a tree node

# @param sum, an integer

# @return a boolean

def pathSum(self, root, sum):

if root == None:

# Empty tree will always result in False

return []

elif root.left == None and root.right == None:

# Reach the leaf.

if root.val == sum: return [[root.val]]

else: return []

elif root.left == None:

# Only has right child.

result = self.pathSum(root.right, sum-root.val)

return [[root.val] + item for item in result]

elif root.right == None:

# Only has left child.

result = self.pathSum(root.left, sum-root.val)

return [[root.val] + item for item in result]

else:

# Has two children.

result = self.pathSum(root.left, sum-root.val)

result.extend(self.pathSum(root.right, sum-root.val))

return [[root.val] + item for item in result]

Solution to Path Sum by LeetCode

6 Jul

Question: https://oj.leetcode.com/problems/path-sum/ Question Name: Path Sum

# Definition for a binary tree node

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

class Solution:

# @param root, a tree node

# @param sum, an integer

# @return a boolean

def hasPathSum(self, root, sum):

if root == None:

# Empty tree will always result in False

return False

elif root.left == None and root.right == None:

# Reach the leaf.

return root.val == sum

elif root.left == None:

# Only has right child.

return self.hasPathSum(root.right, sum-root.val)

elif root.right == None:

# Only has left child.

return self.hasPathSum(root.left, sum-root.val)

else:

# Has two children.

return self.hasPathSum(root.left, sum-root.val) or

self.hasPathSum(root.right, sum-root.val)

Solution to Minimum Depth of Binary Tree by LeetCode

5 Jul

Question: https://oj.leetcode.com/problems/minimum-depth-of-binary-tree/ Question Name: Minimum Depth of Binary Tree

# Definition for a binary tree node

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

class Solution:

# @param root, a tree node

# @return an integer

def minDepth(self, root):

if root == None:

# This is an empty tree.

return 0

elif root.left == None and root.right == None:

# This is a leaf.

return 1

elif root.left == None:

# This non-leaf node only has right son.

return 1 + self.minDepth(root.right)

elif root.right == None:

# This non-leaf node only has left son.

return 1 + self.minDepth(root.left)

else:

# This non-leaf node has both left son and right son.

return 1 + min(self.minDepth(root.left),

self.minDepth(root.right))

Solution to Balanced Binary Tree by LeetCode

5 Jul

Question: https://oj.leetcode.com/problems/balanced-binary-tree/ Question Name: Balanced Binary Tree

# Definition for a binary tree node

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

class Solution:

# @param root, a tree node

# @return a tuple, to say result. result[0] indicates whether the

# subtree, with "current" being the root, is balanced.

# result[1] is -1 if the substree is not balanced, or

# otherwise the height of this subtree.

def isBalancedHelper(self, current):

if current == None:

return (True, 0)

else:

# Check its left subtree

left = self.isBalancedHelper(current.left)

if left[0] == False:

return (False, -1)

# Check its right subtree

right = self.isBalancedHelper(current.right)

if right[0] == False:

return (False, -1)

# Check whether the depth of the two subtrees differ by

# more than 1.

if abs(left[1] - right[1]) <= 1:

height = max(left[1], right[1]) + 1

return (True, height)

else:

return (False, -1)

# @param root, a tree node

# @return a boolean

def isBalanced(self, root):

return self.isBalancedHelper(root)[0]

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