Update solutions.

Author: AnnieKim

BinaryTreeLevelOrderTraversal.h

@@ -42,35 +42,30 @@ class Solution {
         return levelOrder_1(root);
     }
 
-    vector<vector<int>> levelOrder_1(TreeNode *root) {
-        vector<vector<int>> res;
+    vector<vector<int> > levelOrder_1(TreeNode *root) {
+        vector<vector<int> > res;
         if (!root) return res;
-        
         queue<TreeNode *> q;
         q.push(root);
-        q.push(NULL);   // end indicator of one level
+        q.push(NULL);
         vector<int> level;
-        while (q.size() > 0)
+        while (true)
         {
-            TreeNode *front = q.front();
-            q.pop();
-            if (front)
+            TreeNode *node = q.front(); q.pop();
+            if (!node)
             {
-                level.push_back(front->val);
-                if (front->left)
-                    q.push(front->left);
-                if (front->right)
-                    q.push(front->right);
+                res.push_back(level);
+                level.clear();
+                if (q.empty()) break; // end
+                q.push(NULL);
             }
             else
             {
-                if (q.size() > 0)    // CAUTIOUS! infinite loop
-                    q.push(NULL);
-                res.push_back(level);
-                level.clear();
+                level.push_back(node->val);
+                if (node->left) q.push(node->left);
+                if (node->right) q.push(node->right);
             }
         }
-        
         return res;
     }
 

BinaryTreeLevelOrderTraversalII.h

@@ -22,9 +22,7 @@
   [3],
  ]
  
- Solution: 
- 1. Queue version. On the basis of 'Binary Tree Level Order Traversal', reverse the final vector.
- 2. Stack version. Store each level in the vector and then push the vector into stack.
+ Solution: Queue version. On the basis of 'Binary Tree Level Order Traversal', reverse the final vector.
  */
 
 /**
@@ -38,82 +36,32 @@
  */
 class Solution {
 public:
-    vector<vector<int> > levelOrderBottom_1(TreeNode *root) {
+    vector<vector<int> > levelOrderBottom(TreeNode *root) {
         vector<vector<int>> root2leaf;
-        
         queue<TreeNode *> q;
-        if (root)
-        {
-            q.push(root);
-            q.push(NULL);   // end indicator of one level
-        }
-        
+        if (!root) return root2leaf;
+        q.push(root);
+        q.push(NULL);   // end indicator of one level
         vector<int> level;
-        while (q.size() > 0)
+        while (true)
         {
-            TreeNode *front = q.front();
-            q.pop();
-            if (front)
+            TreeNode *node = q.front(); q.pop();
+            if (node)
             {
-                level.push_back(front->val);
-                if (front->left)
-                    q.push(front->left);
-                if (front->right)
-                    q.push(front->right);
+                level.push_back(node->val);
+                if (node->left) q.push(node->left);
+                if (node->right) q.push(node->right);
             }
             else
             {
-                if (q.size() > 0)    // CAUTIOUS! infinite loop
-                    q.push(NULL);
                 root2leaf.push_back(level);
                 level.clear();
+                if (q.empty()) break;    // CAUTIOUS! infinite loop
+                q.push(NULL);
             }
         }
-        
     	// reverse
         reverse(root2leaf.begin(), root2leaf.end());
         return root2leaf;
     }
-
-    vector<vector<int> > levelOrderBottom_2(TreeNode *root) {
-        vector<vector<int>> res;
-        
-        if (!root)
-            return res;
-        
-        stack<vector<TreeNode *>> stk;
-        vector<TreeNode *> level;
-        level.push_back(root);
-        stk.push(level);
-        
-        while (true)
-        {
-            level.clear();
-            for (size_t i = 0; i < stk.top().size(); i++)
-            {
-                if (stk.top()[i]->left)
-                    level.push_back(stk.top()[i]->left);
-                if (stk.top()[i]->right)
-                    level.push_back(stk.top()[i]->right);
-            }
-            
-            if (level.size() == 0)
-                break;
-            
-            stk.push(level);
-        }
-        
-        while (stk.size() > 0)
-        {
-            vector<int> level_val;
-            for (size_t i = 0; i < stk.top().size(); i++)
-            {
-                level_val.push_back(stk.top()[i]->val);
-            }
-            res.push_back(level_val);
-            stk.pop();
-        }
-        
-        return res;
-    }
-};
+};

Combinations.h

@@ -1,6 +1,7 @@
 /*
  Author:     Annie Kim, [email protected]
  Date:       May 5, 2013
+ Update:     Sep 28, 2013
  Problem:    Combinations
  Difficulty: Easy
  Source:     http://leetcode.com/onlinejudge#question_77
@@ -22,25 +23,22 @@
 
 class Solution {
 public:
-    vector<vector<int>> res;
-    vector<vector<int>> combine(int n, int k) {
-        res.clear();
+    vector<vector<int> > combine(int n, int k) {
+        vector<vector<int> > res;
         vector<int> com;
-        combineRe(n, k, com, -1, 0);
+        combineRe(n, k, 1, com, res);
         return res;
     }
-
-    void combineRe(int n, int k, vector<int> &com, int last, int deep)
-    {
-        if (deep == k)
-        {
+    
+    void combineRe(int n, int k, int start, vector<int> &com, vector<vector<int> > &res){
+        int m = com.size();
+        if (m == k) {
             res.push_back(com);
             return;
         }
-        for (int i = last + 1; i < n; ++i)
-        {
-            com.push_back(i+1);
-            combineRe(n, k, com, i, deep + 1);
+        for (int i = start; i <= n-(k-m)+1; ++i) {
+            com.push_back(i);
+            combineRe(n, k, i+1, com, res);
             com.pop_back();
         }
     }

EditDistance.h

@@ -1,7 +1,7 @@
 /*
  Author:     Annie Kim, [email protected]
  Date:       Apr 7, 2013
- Update:     Jul 24, 2013
+ Update:     Sep 28, 2013
  Problem:    Edit Distance
  Difficulty: Medium
  Source:     http://leetcode.com/onlinejudge#question_72
@@ -24,15 +24,12 @@ class Solution {
     }
 
     int minDistance_1(string word1, string word2) {
-        size_t M = word1.size();
-        size_t N = word2.size();
-        
+        int M = word1.size(), N = word2.size();
         int dp[N+1][M+1];
         for (int j = 0; j <= M; j++)
             dp[0][j] = j;
         for (int i = 0; i <= N; i++)
             dp[i][0] = i;
-        
         for (int i = 1; i <= N; i++)
             for (int j = 1; j <= M; j++)
                 if (word2[i-1] == word1[j-1])
@@ -44,33 +41,24 @@ class Solution {
     }
 
     int minDistance_2(string word1, string word2) {
-        int M = word1.size();
-        int N = word2.size();
-        if (M == 0) return N;
-        if (N == 0) return M;
-        
+        int M = word1.size(), N = word2.size();
         int dp[N+1];
         for (int j = 0; j <= N; ++j)
             dp[j] = j;
-        
         for (int i = 1; i <= M; ++i)
         {
-            int left = 0;
-            int cur = 0;
-            for (int j = 0; j <= N; ++j)
+            int upperLeftBackup = dp[0];
+            dp[0] = i;
+            for (int j = 1; j <= N; ++j)
             {
-                if (j == 0)
-                    cur = i;
-                else if (word1[i-1] == word2[j-1])
-                    cur = dp[j-1];
+                int upperLeft = upperLeftBackup;
+                upperLeftBackup = dp[j];
+                if (word1[i-1] == word2[j-1])
+                    dp[j] = upperLeft;
                 else
-                    cur = min(min(dp[j-1], dp[j]), left) + 1;
-                
-                if (j >= 1) dp[j-1] = left;
-                left = cur;
+                    dp[j] = min(min(dp[j-1], dp[j]), upperLeft) + 1;
             }
-            dp[N] = cur;
         }
         return dp[N];
     }
-};
+};

FlattenBinaryTreetoLinkedList.h

@@ -1,6 +1,7 @@
 /*
  Author:     Annie Kim, [email protected]
  Date:       Apr 28, 2013
+ Update:     Sep 29, 2013
  Problem:    Flatten Binary Tree to Linked List
  Difficulty: Medium
  Source:     http://leetcode.com/onlinejudge#question_114
@@ -44,30 +45,20 @@
 class Solution {
 public:
     void flatten(TreeNode *root) {
-        TreeNode *r;
-        flattenRe(root, r);
+        TreeNode *end = NULL;
+        flattenRe(root, end);
     }
-
-    void flattenRe(TreeNode *node, TreeNode *&back)
-    {
-        back = node;
-        if (!node || (!node->left && !node->right)) return;
-
-        TreeNode *leftOrig = node->left;
-        TreeNode *rightOrig = node->right;
-        node->right = leftOrig ? leftOrig : rightOrig;
-        node->left = NULL;
-
-        TreeNode *lr, *rr;
-        flattenRe(leftOrig, lr);
-        flattenRe(rightOrig, rr);
-        back = lr;
-        if (lr) {
-            lr->right = rightOrig;
-        }
-        if (rr) {
-            back = rr;
+    
+    void flattenRe(TreeNode *node, TreeNode *&end) {
+        if (!node) return;
+        TreeNode *lend = NULL, *rend = NULL;
+        flattenRe(node->left, lend);
+        flattenRe(node->right, rend);
+        if (node->left) {
+            lend->right = node->right;
+            node->right = node->left;
+            node->left = NULL;
         }
+        end = rend ? rend : (lend ? lend : node);
     }
-};
-
+};