leetcode Ch4-Binary Tree & BFS & Divide/Conquer

一、

1. Lowest Common Ancestor

 class Solution {
 public:
     TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *A, TreeNode *B) {
         if (root == NULL || root == A || root == B) {
             return root;
         }
         TreeNode* left = lowestCommonAncestor(root->left, A, B);
         TreeNode* right = lowestCommonAncestor(root->right, A, B);
         if (left != NULL && right != NULL) {
             return root;
         }
         if (left != NULL) {
             return left;
         }
         if (right != NULL) {
             return right;
         }
         return NULL;
     }
 };

refer : July，剑指offer

2. Lowest Common Ancestor of a Binary Search Tree

 class Solution {
 public:
     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
         if (root == NULL || p == NULL || q == NULL) {
             return NULL;
         }
         if (root->val > p->val && root->val > q->val) {
             return lowestCommonAncestor(root->left, p, q);
         }
         if (root->val < p->val && root->val < q->val) {
             return lowestCommonAncestor(root->right, p, q);
         }
         return root;
     }
 };

二. Level order [BFS]

1. Binary Tree Level Order Traversal

 class Solution {
 public:
     vector<vector<int>> levelOrder(TreeNode* root) {
         vector<vector<int>> result;
         if (root == NULL) {
             return result;
         }
         queue<TreeNode*> q;
         q.push(root);
         while(!q.empty()) {
             int size = q.size();
             vector<int> v;
             for (int i = ; i < size; i++) {
                 TreeNode* tmp = q.front();
                 q.pop();
                 v.push_back(tmp->val);
                 if (tmp->left != NULL) {
                     q.push(tmp->left);
                 }
                 if (tmp->right != NULL) {
                     q.push(tmp->right);
                 }
             }
             result.push_back(v);
         }
         return result;
     }
 };

2. Binary Tree Level Order Traversal II

 class Solution {
 public:
     vector<vector<int>> levelOrderBottom(TreeNode* root) {
         vector<vector<int>> result;
         if (root == NULL) {
             return result;
         }
         queue<TreeNode*> q;
         q.push(root);
         while(!q.empty()) {
             int size = q.size();
             vector<int> v;
             for (int i = ; i < size; i++) {
                 TreeNode* tmp = q.front();
                 q.pop();
                 v.push_back(tmp->val);
                 if (tmp->left != NULL) {
                     q.push(tmp->left);
                 }
                 if (tmp->right != NULL) {
                     q.push(tmp->right);
                 }
             }
             result.push_back(v);
         }
         reverse(result.begin(), result.end());
         return result;
     }
 };

在1的基础上多加一句reverse即可。

3. Binary Tree Zigzag Level Order Traversal

 class Solution {
 public:
     vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
         vector<vector<int>> result;
         if (root == NULL) {
             return result;
         }
         queue<TreeNode*> q;
         q.push(root);
         int count = ;
         while(!q.empty()) {
             count++;
             int size = q.size();
             vector<int> v;
             for (int i = ; i < size; i++) {
                 TreeNode* tmp = q.front();
                 q.pop();
                 v.push_back(tmp->val);
                 if (tmp->left != NULL) {
                     q.push(tmp->left);
                 }
                 if (tmp->right != NULL) {
                     q.push(tmp->right);
                 }
             }
             if (count %  == ) {
                 reverse(v.begin(), v.end());
             }
             result.push_back(v);
         }
         return result;
     }
 };

在1的基础上多加个count变量，偶数行就reverse一下即可

三、

1. Insert Node in a Binary Search Tree

 TreeNode* insertNode(TreeNode* root, TreeNode* node) {
     if (root == NULL) {
         return node;
     }
     if (node->val > root->val) {
         root->right = insertNode(root->right, node);
     } else {
         root->left = insertNode(root->left, node);
     }
     return root;
 }

2. Search Range in Binary Search Tree

code1:

 class Solution {
 public:
     vector<int> searchRange(TreeNode* root, int k1, int k2) {
         helper(root, k1, k2);
         return result;
     }
     void helper(TreeNode* root, int k1, int k2) {
         if (root == NULL) {
             return;
         }
         if (k1 < root->val) {//说明左子树里有可能有
             helper(root->left, k1, k2);
         }
         if (root->val >= k1 && root->val <= k2) {
             result.push_back(root->val);
         }
         if (k2 > root->val) {
             helper(root->right, k1, k2);
         }
     }
 private:
     vector<int>    result;
 };

code2: 自己实现的，太繁琐。

 vector<int> searchRange(TreeNode* root, int k1, int k2) {
     vector<int> result;
     if (root == NULL) {
         return result;
     }
     if (root->val < k1) {
         return searchRange(root->right, k1, k2);
     }
     if (root->val > k2) {
         return searchRange(root->left, k1, k2);
     }
     if (root->val >= k1 && root->val <= k2) {
         vector<int> tmp1 = searchRange(root->left, k1, root->val - );
         vector<int> tmp2 = searchRange(root->right, root->val + , k2);
         result.insert(result.end(), tmp1.begin(), tmp1.end());
         result.push_back(root->val);
         result.insert(result.end(), tmp2.begin(), tmp2.end());
     }
     return result;
 }

Binary Search Tree Iterator

 class BSTIterator {
 public:
     BSTIterator(TreeNode* root) {
         pushAll(root);
     }

     bool hasNext() {
         return (!myStack.empty());
     }

     int next() {
         TreeNode* tmp = myStack.top();
         myStack.pop();
         pushAll(tmp->right);
         return tmp->val;
     }

 private:
     stack<TreeNode*> myStack;
     void pushAll(TreeNode* node);
 };

 void BSTIterator::pushAll(TreeNode* node) {
     while (node != NULL) {
         myStack.push(node);
         node = node->left;
     }
 }

 /**
  * Your BSTIterator will be called like this:
  * BSTIterator i = BSTIterator(root);
  * while (i.hasNext()) cout << i.next();
  */

Remove Node in Binary Search Tree

Binary Tree Maximum Path Sum

参见 ref十五

Binary Tree Serialization

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对于n个数的数组，一个数x如果从左往右数是第k个数，那么从右往左数的话是第(n - k + 1)个数。