107. Binary Tree Level Order Traversal II
题目
直接代码：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int>> levelOrderBottom(TreeNode* root) { 13 vector<vector<int>> res; 14 vector<int> temp; 15 if(NULL == root) 16 return res; 17 18 queue<TreeNode*> myQue; 19 TreeNode *temp1; 20 bool flag = true; 21 myQue.push(root); 22 23 while (!myQue.empty()) 24 { 25 temp.clear(); 26 myQue.push(NULL); 27 temp1 = myQue.front(); 28 myQue.pop(); 29 while (NULL != temp1) 30 { 31 temp.push_back(temp1->val); 32 if(NULL != temp1->left) 33 myQue.push(temp1->left); 34 if(NULL != temp1->right) 35 myQue.push(temp1->right); 36 temp1 = myQue.front(); 37 myQue.pop(); 38 } 39 40 41 res.push_back(temp); 42 } 43 reverse(res.begin(),res.end()); 44 return res; 45 } 46 47 };
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108. Convert Sorted Array to Binary Search Tree
题目
分析：
讲一个排好序的数组，构造成平衡二叉树，并且是二叉搜索树，其基本思想是折半（二分）法。
代码如下：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 TreeNode* sortedArrayToBST(vector<int>& nums) { 13 int size = nums.size(); 14 if(0 == size) 15 return NULL; 16 TreeNode *root = mySortedArrayToBST(nums,0,size-1); 17 return root; 18 } 19 TreeNode* mySortedArrayToBST(vector<int>& nums,int start,int end) 20 { 21 int middle = (start+end)/2; 22 TreeNode *temp = new TreeNode(nums[middle]); 23 if(middle == start) 24 temp->left = NULL; 25 else 26 temp->left = mySortedArrayToBST(nums,start,middle-1); 27 28 if(middle == end) 29 temp->right = NULL; 30 else 31 temp->right = mySortedArrayToBST(nums,middle+1,end); 32 33 return temp; 34 } 35 };