LeetCode – Page 10

Solution to Maximum Depth of Binary Tree by LeetCode

4 Jul

Question: https://oj.leetcode.com/problems/maximum-depth-of-binary-tree/ Question Name: Maximum 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 maxDepth(self, root):

if root == None:

return 0

else:

return 1 + max(self.maxDepth(root.left),

self.maxDepth(root.right))

Solution to Binary Tree Zigzag Level Order Traversal by LeetCode

4 Jul

Question: https://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ Question Name: Binary Tree Zigzag Level Order Traversal

# 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 list of lists of integers

def zigzagLevelOrder(self, root):

if root == None: return []

# In the first stage, we store hte nodes (not their value).

result = []

nextLayer = [root]

left2Right = True # The direction to travel in current layer.

while len(nextLayer) != 0:

result.append(nextLayer)

# Gather the nodes in the next deeper layer.

nextLayer = []

left2Right = not left2Right

# Access the fathers in previous layer in manner of stack. So

# everytime the traversal order is reversed.

index = len(result[-1]) - 1

while index > -1:

father = result[-1][index]

if left2Right:

if father.left != None: nextLayer.append(father.left)

if father.right != None: nextLayer.append(father.right)

else:

if father.right != None: nextLayer.append(father.right)

if father.left != None: nextLayer.append(father.left)

index -= 1

# In the second stage, we convert the nodes into their values.

result = [[node.val for node in layer] for layer in result]

return result

Solution to Binary Tree Level Order Traversal by LeetCode

4 Jul

Question: https://oj.leetcode.com/problems/binary-tree-level-order-traversal/ Question Name: Binary Tree Level Order Traversal

# 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 list of lists of integers

def levelOrder(self, root):

if root == None: return []

# In the first stage, we store hte nodes (not their value).

result = []

nextLayer = [root]

while len(nextLayer) != 0:

result.append(nextLayer)

# Gather the nodes in the next deeper layer.

nextLayer = []

for father in result[-1]:

if father.left != None: nextLayer.append(father.left)

if father.right != None: nextLayer.append(father.right)

# In the second stage, we convert the nodes into their values.

result = [[node.val for node in layer] for layer in result]

return result

Solution to Symmetric Tree by LeetCode

3 Jul

Question: https://oj.leetcode.com/problems/symmetric-tree/ Question Name: Symmetric Tree

# Definition for a binary tree node

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

class Solution:

def isSymmetricHelp(self, left, right):

if left == None and right == None:

return True

elif left == None or right == None:

return False

elif left.val != right.val:

return False

else:

return self.isSymmetricHelp(left.right, right.left) and

self.isSymmetricHelp(left.left, right.right)

# @param root, a tree node

# @return a boolean

def isSymmetric(self, root):

if root == None: return True

else: return self.isSymmetricHelp(root.left, root.right)

Solution to Recover Binary Search Tree by LeetCode

3 Jul

Question: https://oj.leetcode.com/problems/recover-binary-search-tree/ Question Name: Recover Binary Search 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 tree node

def recoverTree(self, root):

''' The solution is similar with the in-order travel. We

need to find out the rule-breaker, whose value is

greater than its successor. Then we need to find out

the last node, whose value is smaller that the rule-

breaker's. Finanly, we will exchange the value of

these two nodes.

'''

if root == None: return root

stack = [root]

previous = None

current = root.left

toExchangeLeft, toExchangeRight = None, None

while len(stack) != 0 or current != None:

while current != None:

stack.append(current)

current = current.left

current = stack.pop()

if previous != None:

if toExchangeLeft == None:

# Find the rule-breaker, whose value is greater

# then its successor.

if previous.val > current.val:

toExchangeLeft = previous

else:

# Find the last node, whose value is smaller

# than the rule-breaker.

if current.val > toExchangeLeft.val:

toExchangeRight = previous

break

previous = current

current = current.right

# We checked the all the nodes except the last one. If we did not

# find one to be exchanged with the rule-break, the last node

# will be used.

if toExchangeRight == None: toExchangeRight = previous

toExchangeLeft.val, toExchangeRight.val = toExchangeRight.val, toExchangeLeft.val

return root

Solution to Same Tree by LeetCode

2 Jul

Question: https://oj.leetcode.com/problems/same-tree/ Question Name: Same 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 p, a tree node

# @param q, a tree node

# @return a boolean

def isSameTree(self, p, q):

if p == None and q == None:

return True

elif p == None or q == None:

return False

elif p.val == q.val:

return self.isSameTree(p.left, q.left) and

self.isSameTree(p.right, q.right)

else:

return False

Solution to Validate Binary Search Tree by LeetCode

2 Jul

Question: https://oj.leetcode.com/problems/validate-binary-search-tree/ Question Name: Validate Binary Search Tree This is an extension of an older question Binary Tree Inorder Traversal.

# 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 list of integers

def inorderTraversal(self, root):

# Handle with the special cast (empty input)

if root == None: return []

stack = [root]

current = root.left

result = []

while len(stack) != 0 or current != None:

# Find the leftmost node among all the un-accessed ones.

while current != None:

stack.append(current)

current = current.left

# Fetch the value in currently leftmost node

current = stack.pop()

result.append(current.val)

# Try the right son of current node

current = current.right

return result

# @param root, a tree node

# @return a boolean

def isValidBST(self, root):

inOrder = self.inorderTraversal(root)

# If root is a binary search tree, the result of in order

# travle will be strickly increasing.

for index in xrange(len(inOrder)-1):

if inOrder[index] >= inOrder[index+1]:

return False

return True

Solution to Interleaving String by LeetCode

2 Jul

Question: https://oj.leetcode.com/problems/interleaving-string/ Question Name: Interleaving String

class Solution:

# @return a boolean

def isInterleaveHelper(self, s1, s1Index, s2, s2Index, s3, s3Index,

accessed):

if (s1Index, s2Index) in accessed:

# Have tried. Did not success. Wont success.

return False

else:

accessed.add((s1Index, s2Index))

# Checked all contents in s3. s3 is formed by the interleaving of

# s1 and s2

if s3Index == len(s3):

return True

# Try to use the next character in s1

if s1Index != len(s1) and s1[s1Index] == s3[s3Index]:

if self.isInterleaveHelper(s1, s1Index+1, s2, s2Index, s3,

s3Index+1, accessed):

return True

# Try to use the next character in s2

if s2Index != len(s2) and s2[s2Index] == s3[s3Index]:

if self.isInterleaveHelper(s1, s1Index, s2, s2Index+1, s3,

s3Index+1, accessed):

return True

return False

# @return a boolean

def isInterleave(self, s1, s2, s3):

# Check length firstly.

if len(s1) + len(s2) != len(s3):

return False

# Then check the character count.

charDis = {}

for item in s1: charDis[item] = charDis.get(item, 0) + 1

for item in s2: charDis[item] = charDis.get(item, 0) + 1

for item in s3: charDis[item] = charDis.get(item, 0) - 1

for item in charDis:

if charDis[item] != 0: return False

# Finally check overleaving

return self.isInterleaveHelper(s1, 0, s2, 0, s3, 0, set())

Solution to Binary Tree Inorder Traversal by LeetCode

1 Jul

Question: https://oj.leetcode.com/problems/binary-tree-inorder-traversal/ Question Name: Binary Tree Inorder Traversal

# 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 list of integers

def inorderTraversal(self, root):

# Handle with the special cast (empty input)

if root == None: return []

stack = [root]

current = root.left

result = []

while len(stack) != 0 or current != None:

# Find the leftmost node among all the un-accessed ones.

while current != None:

stack.append(current)

current = current.left

# Fetch the value in currently leftmost node

current = stack.pop()

result.append(current.val)

# Try the right son of current node

current = current.right

return result

Solution to Unique Binary Search Trees II by LeetCode

1 Jul

Question: https://oj.leetcode.com/problems/unique-binary-search-trees-ii/ Question Name: Unique Binary Search Trees 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 a list(representing trees)

# @return the root node the tree

def convertArray2Trees(self, array):

result = [TreeNode(i+1) for i in xrange(len(array))]

root = None

for index in xrange(len(array)):

if array[index] == -1:

# This node has no father. It is the root node.

root = result[index]

else:

if array[index] > index:

# Its father is on the right side. Thus it is

# father's left son.

result[array[index]].left = result[index]

else:

# Its father is no the left side.

result[array[index]].right = result[index]

return root

# @return a list of lists(representing trees)

def generateArray(self, array, current, left, right):

result = []

resultTemp = []

# Not possible to have any son.

if left >= current and current >= right:

return [array]

# Fill the left son

if left >= current:

resultTemp = [array]

else:

for son in xrange(left, current):

temp = array * 1

temp[son] = current

resultTemp.extend(self.generateArray(temp, son, left, current-1))

# Fill the right son

if current >= right:

result = resultTemp

else:

for tempStatus in resultTemp:

for son in xrange(current+1, right+1):

temp = tempStatus * 1

temp[son] = current

result.extend(self.generateArray(temp, son, current+1, right))

return result

# @return a list of tree node

def generateTrees(self, n):

# Handle the special case.

if n == 0: return [None]

result = []

# We firstly generate the trees in form of array.

# array[i] == j means, the i(th) node's father is j(th) node. (0-based)

for root in xrange(n):

temp = [-1] * n

tempRes = self.generateArray(temp, root, 0, n-1)

result.extend(tempRes)

# Convert the arrays into trees.

return [self.convertArray2Trees(array) for array in result]

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