1+ using System ;
2+ using System . Collections . Generic ;
3+ using System . Linq ;
4+ using System . Text ;
5+ using System . Threading . Tasks ;
6+ 7+ namespace LowestCommonAncestorBTree
8+ {
9+ /// <summary>
10+ /// code review on July 9, 2018
11+ /// Leetcode 236
12+ /// https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
13+ /// </summary>
14+ class TreeNode
15+ {
16+ public int val ;
17+ public TreeNode left ;
18+ public TreeNode right ;
19+ 20+ public TreeNode ( int x )
21+ {
22+ left = null ;
23+ right = null ;
24+ val = x ;
25+ }
26+ } ;
27+ 28+ class LowestCommonAncestorB
29+ {
30+ /*
31+ * Leetcode:
32+ * Lowest common ancestor in binary tree
33+ *
34+ * Try a few of implementations - 4 different ones:
35+ * 1. Method A:
36+ * Top down method, using counting method to help
37+ * worst case time O(N^2)
38+ *
39+ * 2. Method B:
40+ * A Bottom-up Approach (Worst case O(n) ):
41+ Using a bottom-up approach, we can improve over the top-down approach
42+ by avoiding traversing the same nodes over and over again.
43+ */
44+ static void Main ( string [ ] args )
45+ {
46+ var root = new TreeNode ( 1 ) ;
47+ root . left = new TreeNode ( 2 ) ;
48+ root . right = new TreeNode ( 3 ) ;
49+ 50+ root . left . left = new TreeNode ( 4 ) ;
51+ root . left . right = new TreeNode ( 5 ) ;
52+ 53+ root . right . left = new TreeNode ( 6 ) ;
54+ root . right . right = new TreeNode ( 7 ) ;
55+ 56+ var ancestor = LowestCommonAncestor ( root , root . right . left , root . left . right ) ;
57+ }
58+ 59+ /*
60+ * source code from blog:
61+ * http://articles.leetcode.com/2011/07/lowest-common-ancestor-of-a-binary-tree-part-i.html
62+ *
63+ * A Top-Down Approach (Worst case O(n^2) ):
64+ *
65+ * julia's comment:
66+ 1. check left subtree count of matching p or q:
67+ 0 - no match in left sub tree, so go to check right subtree
68+ 1 - found it, the root is the node
69+ 2 - continue to left, check
70+ * 2. put the code in the certain order: root, left, right
71+ * so, making it easy to follow.
72+ * two orders:
73+ * 1. check countingProblem return value, from 0 to 1 to 2, increasing order
74+ * 2. using countingProblem calculation take action, "found it", "go left",
75+ * "go right" three choice.
76+ *
77+ * Keep the code in the above order, this way, easy to memorize, explain, and discuss.
78+ *
79+ * 3. Using a easy counting problem to help solve lowest common ancestor (LCA),
80+ * What is the thinking process?
81+ * Here is the scripts Julia makes out:
82+ * Go to visit root first, check it value, null or one of two nodes,
83+ * and then, decide to find LCA, or go left to find it or go right to find it.
84+ *
85+ * But wait, how to decide to find it, or go left/ right, so calculate the left sub tree
86+ * matching count first, use its value to determine.
87+ *
88+ * 4. online judge: (lowest common ancestor in binary search tree, not in binary tree)
89+ * 27 / 27 test cases passed.
90+ Status: Accepted
91+ Runtime: 180 ms
92+ *
93+ */
94+ public static TreeNode LowestCommonAncestor ( TreeNode root , TreeNode p , TreeNode q )
95+ {
96+ if ( root == null || p == null || q == null )
97+ return null ;
98+ 99+ if ( root == p || root == q )
100+ return root ;
101+ 102+ // go left
103+ int leftTreeMatched = findMatches ( root . left , p , q ) ;
104+ 105+ bool zeroFound = leftTreeMatched == 0 ;
106+ bool oneFound = leftTreeMatched == 1 ;
107+ bool twoFound = leftTreeMatched == 2 ;
108+ 109+ if ( oneFound )
110+ {
111+ return root ;
112+ }
113+ else if ( twoFound )
114+ {
115+ return LowestCommonAncestor ( root . left , p , q ) ;
116+ }
117+ else if ( zeroFound )
118+ {
119+ return LowestCommonAncestor ( root . right , p , q ) ;
120+ }
121+ 122+ return null ;
123+ }
124+ 125+ /*
126+ * source code from blog:
127+ * http://articles.leetcode.com/2011/07/lowest-common-ancestor-of-a-binary-tree-part-i.html
128+ * comment from blog:
129+ * A Top-Down Approach (Worst case O(n^2) ):
130+
131+ Let’s try the top-down approach where we traverse the nodes from the top to the bottom.
132+ * First, if the current node is one of the two nodes, it must be the LCA of the two nodes.
133+ * If not, we count the number of nodes that matches either p or q in the left subtree (which
134+ * we call totalMatches). If totalMatches equals 1, then we know the right subtree will contain
135+ * the other node. Therefore, the current node must be the LCA. If totalMatches equals 2, we know
136+ * that both nodes are contained in the left subtree, so we traverse to its left child. Similar
137+ * with the case where totalMatches equals 0 where we traverse to its right child.
138+ *
139+ * Return #nodes that matches P or Q in the subtree.
140+ *
141+ * julia's comment:
142+ * 1. Convert C++ code to C#
143+ * 2. Solve a problem first: counting matches in the tree
144+ * Counting always helps to make decision. Using number to make a decision
145+ * 3. think about the question:
146+ * Can this function only uses two line code?
147+ *
148+ *
149+ */
150+ private static int findMatches ( TreeNode root , TreeNode p , TreeNode q )
151+ {
152+ if ( root == null ) return 0 ;
153+ 154+ int matches = findMatches ( root . left , p , q ) + findMatches ( root . right , p , q ) ;
155+ 156+ if ( root == p || root == q )
157+ {
158+ return 1 + matches ;
159+ }
160+ else
161+ {
162+ return matches ;
163+ }
164+ }
165+ }
166+ }
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