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二分 Convert Sorted Array to Binary Search Tree

2015-06-01 14:23 337 查看
思路:

递归+二分法。(因为是height balanced,即高度平衡的二叉树)

/**
 * 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 {
private:
    template<typename RandomAccessIterator>
    TreeNode* buildBST(RandomAccessIterator first, RandomAccessIterator last) {
        const auto length = distance(first, last);
        if(length <= 0) return nullptr;
        auto mid = first + length/2;
        TreeNode *root = new TreeNode(*mid);
        root->left = buildBST(first, mid);
        root->right = buildBST(mid+1, last);
        return root;
    }
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return buildBST(nums.begin(), nums.end());
    }
};
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