Sheng – Page 17 – Code Says

Solution to Convert Sorted List to Binary Search Tree by LeetCode

5 Jul

Question: https://oj.leetcode.com/problems/convert-sorted-list-to-binary-search-tree/ Question Name: Convert Sorted List to Binary Search Tree

# Definition for a binary tree node

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

# Definition for singly-linked list.

# class ListNode:

# def __init__(self, x):

# self.val = x

# self.next = None

class Solution:

# @param num, a list of integers

# @return a tree node

def sortedArrayToBST(self, num):

''' With a given array num, the root of the corresponding tree

is always the middle one in the array.

'''

treeSize = len(num)

if treeSize == 0:

return None

else:

root = TreeNode(num[treeSize//2])

root.left = self.sortedArrayToBST(num[:treeSize//2])

root.right = self.sortedArrayToBST(num[treeSize//2+1:])

return root

# @param head, a list node

# @return a tree node

def sortedListToBST(self, head):

''' Firstly we convert the list into an array.

Then we construct the BST with that array.

'''

num = []

current = head

while current != None:

num.append(current.val)

current = current.next

return self.sortedArrayToBST(num)

Solution to Convert Sorted Array to Binary Search Tree by LeetCode

5 Jul

Question: https://oj.leetcode.com/problems/convert-sorted-array-to-binary-search-tree/ Question Name: Convert Sorted Array to 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 num, a list of integers

# @return a tree node

def sortedArrayToBST(self, num):

''' With a given array num, the root of the corresponding tree

is always the middle one in the array.

'''

treeSize = len(num)

if treeSize == 0:

return None

else:

root = TreeNode(num[treeSize//2])

root.left = self.sortedArrayToBST(num[:treeSize//2])

root.right = self.sortedArrayToBST(num[treeSize//2+1:])

return root

Solution to Binary Tree Level Order Traversal II by LeetCode

5 Jul

Question: https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/ Question Name: Binary Tree Level Order Traversal II Nearly the same as the Binary Tree Level Order Traversal. We only need to reverse the output.

# 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 levelOrderBottom(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[::-1]

Solution to Construct Binary Tree from Preorder and Inorder Traversal by LeetCode

5 Jul

Question: https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ Question Name: Construct Binary Tree from Preorder and 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 inorder, a list of integers

# @param inOrderZone, an array to limit the nodes for current subtree

# inOrderZone[0] is the beginning position of current subtree

# in inorder array. inOrderZone[1] is the end position.

# @param inPos, a dictionary with the position of each char in inorder.

# @param preorder, a list of integers

# @param preOrderZone, an array to limit the nodes for current subtree.

# Same as inOrderZone.

# @param prePos, a dictionary with the position of each char in preorder.

# @return a tree node

def buildTreeHelper(self, inorder, inOrderZone, inPos,

preorder, preOrderZone, prePos):

''' Built the tree recuisively. The first element in preOrderZone

will always be the root of current tree (or subtree).

'''

# Reach the leaf.

if inOrderZone[1] < inOrderZone[0] : return None

# Create the root node of current (sub)tree

# The root node is in some middle place among inOrderZone.

# The root node is in the first position of current preOrderZone.

root = TreeNode(preorder[preOrderZone[0]])

# Recursively build the left son of current root

leftInOrderZone = [inOrderZone[0], inPos[root.val]-1]

leftCount = inPos[root.val] - inOrderZone[0]

leftPreOrderZone = [preOrderZone[0]+1, preOrderZone[0]+leftCount]

root.left = self.buildTreeHelper(inorder, leftInOrderZone, inPos,

preorder, leftPreOrderZone, prePos)

# Recursively build the right son of current root

rightInOrderZone = [inPos[root.val]+1, inOrderZone[1]]

rightCount = inOrderZone[1] - inPos[root.val]

rightPreOrderZone = [preOrderZone[1]-rightCount+1, preOrderZone[1]]

root.right = self.buildTreeHelper(inorder, rightInOrderZone, inPos,

preorder, rightPreOrderZone, prePos)

return root

# @param inorder, a list of integers

# @param preorder, a list of integers

# @return a tree node

def buildTree(self, preorder, inorder):

# There is not duplicate in the tree. And we record the position for

# each charater in the inorder array and preorder array.

inPos = {}

prePos = {}

for index in xrange(len(inorder)):

inPos[inorder[index]] = index

prePos[preorder[index]] = index

return self.buildTreeHelper(inorder, [0, len(inorder)-1], inPos,

preorder, [0, len(preorder)-1], prePos)

Solution to Construct Binary Tree from Inorder and Postorder Traversal by LeetCode

5 Jul

Question: https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ Question Name: Construct Binary Tree from Inorder and Postorder 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 inorder, a list of integers

# @param inOrderZone, an array to limit the nodes for current subtree

# inOrderZone[0] is the beginning position of current subtree

# in inorder array. inOrderZone[1] is the end position.

# @param inPos, a dictionary with the position of each char in inorder.

# @param postorder, a list of integers

# @param postOrderZone, an array to limit the nodes for current subtree.

# Same as inOrderZone.

# @param postPos, a dictionary with the position of each char in postorder.

# @return a tree node

def buildTreeHelper(self, inorder, inOrderZone, inPos,

postorder, postOrderZone, postPos):

''' Built the tree recuisively. The last element in postOrderZone

array will always be the root of current tree (or subtree).

'''

# Reach the leaf.

if inOrderZone[1] < inOrderZone[0] : return None

# Create the root node of current (sub)tree

# The root node is in some middle place among inOrderZone.

# The root node is in the last position of current postOrderZone.

root = TreeNode(postorder[postOrderZone[1]])

# Recursively build the left son of current root

leftInOrderZone = [inOrderZone[0], inPos[root.val]-1]

leftCount = inPos[root.val] - inOrderZone[0]

leftPostOrderZone = [postOrderZone[0], postOrderZone[0]+leftCount-1]

root.left = self.buildTreeHelper(inorder, leftInOrderZone, inPos,

postorder, leftPostOrderZone, postPos)

# Recursively build the right son of current root

rightInOrderZone = [inPos[root.val]+1, inOrderZone[1]]

rightCount = inOrderZone[1] - inPos[root.val]

rightPostOrderZone = [postOrderZone[1]-rightCount, postOrderZone[1]-1]

root.right = self.buildTreeHelper(inorder, rightInOrderZone, inPos,

postorder, rightPostOrderZone, postPos)

return root

# @param inorder, a list of integers

# @param postorder, a list of integers

# @return a tree node

def buildTree(self, inorder, postorder):

# There is not duplicate in the tree. And we record the position for

# each charater in the inorder array and postorder array.

inPos = {}

postPos = {}

for index in xrange(len(inorder)):

inPos[inorder[index]] = index

postPos[postorder[index]] = index

return self.buildTreeHelper(inorder, [0, len(inorder)-1], inPos,

postorder, [0, len(postorder)-1], postPos)

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

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