Adding solution of 116, 117, 118, 119 problems

Author: Garvit244

100-200q/116.py

@@ -0,0 +1,58 @@
+'''
+	Given a binary tree
+
+	struct TreeLinkNode {
+	  TreeLinkNode *left;
+	  TreeLinkNode *right;
+	  TreeLinkNode *next;
+	}
+	Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
+
+	Initially, all next pointers are set to NULL
+	Example:
+
+	Given the following perfect binary tree,
+
+	     1
+	   /  \
+	  2    3
+	 / \  / \
+	4  5  6  7
+	After calling your function, the tree should look like:
+
+	     1 -> NULL
+	   /  \
+	  2 -> 3 -> NULL
+	 / \  / \
+	4->5->6->7 -> NULL
+'''
+
+# Definition for binary tree with next pointer.
+# class TreeLinkNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+#         self.next = None
+
+class Solution:
+    # @param root, a tree link node
+    # @return nothing
+    def connect(self, root):
+        def recursive(node):
+        	if node is None:
+        		return
+
+        	if node.left:
+        		node.left.next = node.right
+        	if node.right:
+        		if node.next:
+	        		node.right.next = node.next.left
+	        	else:
+	        		node.right.next  = None
+	        recursive(node.left)
+	        recursive(node.right)
+
+		if root != None:
+			root.next = None
+			recursive(root)

100-200q/117.py

@@ -0,0 +1,57 @@
+'''
+	Given a binary tree
+
+	struct TreeLinkNode {
+	  TreeLinkNode *left;
+	  TreeLinkNode *right;
+	  TreeLinkNode *next;
+	}
+	Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
+
+	Initially, all next pointers are set to NULL.
+
+	Example:
+
+	Given the following binary tree,
+
+	     1
+	   /  \
+	  2    3
+	 / \    \
+	4   5    7
+	After calling your function, the tree should look like:
+
+	     1 -> NULL
+	   /  \
+	  2 -> 3 -> NULL
+	 / \    \
+	4-> 5 -> 7 -> NULL
+'''
+
+# Definition for binary tree with next pointer.
+# class TreeLinkNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+#         self.next = None
+
+class Solution:
+    # @param root, a tree link node
+    # @return nothing
+    def connect(self, root):
+    	if root == None:
+    		return 
+    	queue = [root]
+    	queue.append(None)
+
+    	while queue:
+    		front = queue.pop(0)
+    		if front is not None:
+    			front.next = queue[0]
+    			if front.left:
+    				queue.append(front.left)
+    			if front.right:
+    				queue.append(front.right)
+    		elif queue:
+    			queue.append(None)

100-200q/118.py

@@ -0,0 +1,31 @@
+'''
+	Given a non-negative integer numRows, generate the first numRows of Pascal's triangle.
+
+	Example:
+
+	Input: 5
+	Output:
+	[
+	     [1],
+	    [1,1],
+	   [1,2,1],
+	  [1,3,3,1],
+	 [1,4,6,4,1]
+	]
+'''
+class Solution(object):
+    def generate(self, numRows):
+        """
+        :type numRows: int
+        :rtype: List[List[int]]
+        """
+        triangle = []
+
+        for row in range(numRows):
+        	new_row = [0 for _ in range(row+1)]
+        	new_row[0], new_row[-1] = 1, 1
+
+        	for col in range(1, len(new_row)-1):
+        		new_row[col] = triangle[row-1][col-1] + triangle[row-1][col]
+        	triangle.append(new_row)
+        return triangle

100-200q/119.py

@@ -0,0 +1,16 @@
+'''
+	Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal's triangle.
+
+	Note that the row index starts from 0.
+'''
+class Solution(object):
+    def getRow(self, rowIndex):
+        """
+        :type rowIndex: int
+        :rtype: List[int]
+        """
+        row = [1]*(rowIndex+1)
+        for i in range(1, rowIndex+1):
+        	for j in range(i-1, 0, -1):
+        		row[j] += row[j-1]
+        return row