[leetcode]convert-sorted-array-to-binary-search-tree

RT

思路

a height balanced BST
BST的中序遍历是一个sorted-array，再构造回去成一个BST,先将中间的元素作为根节点，这个节点的左右分别是左子树和右子树。如此递归地进行即可。

Solution 1

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 class Solution {
 public:
     TreeNode *sortedArrayToBST(vector<int> &num) {
         if (num.size() == 0){
             return nullptr;
         }
         TreeNode* root = new TreeNode(0);
         sortedArray2BST(num, root); //不被初始化的指针不能用来赋值，编译不通过
         return root;
     }

     void sortedArray2BST(vector<int> &num, TreeNode* node){
         //cout<<num.size()<<endl;

         int mid = num.size() / 2;
         int val = num[mid];
         vector<int>::iterator it_begin = num.begin();
         vector<int>::iterator it_end = num.end();
         node->val = val;
         //cout<<"val:"<<val<<endl;
         if (mid > 0){ //防止切分之后的vec为空
             vector<int> left_vec(it_begin, it_begin + mid); //attention the index
             node->left = new TreeNode(0);
             sortedArray2BST(left_vec, node->left);
         }
         if (mid != num.size() - 1){ //防止切分之后的vec为空
             vector<int> right_vec(it_begin + mid+1, it_end); //attention the index
             node->right = new TreeNode(0);
             sortedArray2BST(right_vec, node->right);
         }
     }

    ~Solution(){
        //delete the tree nodes
    }
 };