New Problem "Lowest Common Ancestor of a Binary Tree"

haoel · haoel · commit afd3a244be7c · 2015-07-17T09:29:56.000+08:00
diff --git a/README.md b/README.md
@@ -7,7 +7,8 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
-|218|[Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)| [C++](./algorithms/lowestCommonAncestorOfABinarySearchTree/LowestCommonAncestorOfABinarySearchTree.cpp)|Easy|
+|220|[Lowest Common Ancestor of a Binary Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/)| [C++](./algorithms/lowestCommonAncestorOfABinaryTree/LowestCommonAncestorOfABinaryTree.cpp)|Easy|
+|219|[Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)| [C++](./algorithms/lowestCommonAncestorOfABinarySearchTree/LowestCommonAncestorOfABinarySearchTree.cpp)|Easy|
 |218|[Palindrome Linked List](https://leetcode.com/problems/palindrome-linked-list/)| [C++](./algorithms/palindromeLinkedList/PalindromeLinkedList.cpp)|Easy|
 |217|[Number of Digit One](https://leetcode.com/problems/number-of-digit-one/)| [C++](./algorithms/numberOfDigitOne/NumberOfDigitOne.cpp)|Medium|
 |216|[Implement Queue using Stacks](https://leetcode.com/problems/implement-queue-using-stacks/)| [C++](./algorithms/implementQueueUsingStacks/ImplementQueueUsingStacks.cpp)|Easy|
diff --git a/algorithms/lowestCommonAncestorOfABinaryTree/LowestCommonAncestorOfABinaryTree.cpp b/algorithms/lowestCommonAncestorOfABinaryTree/LowestCommonAncestorOfABinaryTree.cpp
@@ -0,0 +1,102 @@
+// Source : https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/
+// Author : Hao Chen
+// Date   : 2015-07-17
+
+/********************************************************************************** 
+ * 
+ * Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the 
+ * tree.
+ * 
+ * According to the definition of LCA on Wikipedia: “The lowest common ancestor is 
+ * defined between two nodes v and w as the lowest node in T that has both v and w as 
+ * descendants (where we allow a node to be a descendant of itself).”
+ * 
+ *         _______3______
+ *        /              \
+ *     ___5__          ___1__
+ *    /      \        /      \
+ *    6      _2       0       8
+ *          /  \
+ *          7   4
+ * 
+ * For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example 
+ * is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according 
+ * to the LCA definition.
+ *               
+ *               
+ *               
+ **********************************************************************************/
+
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    bool findPath(TreeNode* root, TreeNode* p, vector<TreeNode*>& path) {
+        if (root==NULL) return false;
+        if (root == p) {
+            path.push_back(p);
+            return true;
+        }
+        
+        path.push_back(root);
+        if (findPath(root->left, p, path)) return true;
+        if (findPath(root->right, p, path)) return true;
+        path.pop_back();
+        
+        return false;
+    }
+
+    //Ordinary way, find the path and comapre the path.
+    TreeNode* lowestCommonAncestor01(TreeNode* root, TreeNode* p, TreeNode* q) {
+        
+        vector<TreeNode*> path1, path2;
+        
+        if (!findPath(root, p, path1)) return NULL;
+        if (!findPath(root, q, path2)) return NULL;
+        
+        int len = path1.size() < path2.size() ? path1.size() : path2.size();
+        TreeNode* result = root;
+        for(int i=0; i<len; i++) {
+            if (path1[i] != path2[i]) {
+                return result;
+            }
+            result = path1[i];
+        }
+        return result;
+    }
+    
+    //Actually, we can do the recursive search in one time
+    TreeNode* lowestCommonAncestor02(TreeNode* root, TreeNode* p, TreeNode* q) {
+        
+        //return if found or not found, return NULL if not found
+        if (root==NULL || root == p || root == q) return root;
+        
+        //find the LCA in left tree
+        TreeNode* left = lowestCommonAncestor02(root->left, p, q);
+        //find the LCA in right tree
+        TreeNode* right = lowestCommonAncestor02(root->right, p, q);
+        
+        //left==NULL means both `p` and `q` are not found in left tree.
+        if (left==NULL) return right;
+        //right==NULL means both `p` and `q` are not found in right tree.
+        if (right==NULL) return left;
+        // left!=NULL && right !=NULL, which means `p` & `q` are seperated in left and right tree.
+        return root;
+    }
+    
+    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
+        srand(time(0));
+        if (random()%2) {
+            return lowestCommonAncestor02(root, p, q);
+        }
+        return lowestCommonAncestor01(root, p, q);
+    }
+};
