Tony's Log

Algorithms, Distributed System, Machine Learning

  博客园 :: 首页 :: 博问 :: 闪存 :: 新随笔 :: 联系 :: 订阅 订阅 :: 管理 ::

Another textbook problem. We take back of postorder array as the current root, and then we can split the inorder array: 1st half for current right child, and 2nd for current left.

class Solution {
public:
    TreeNode *_buildTree(int in[], int i0, int i1, int insize, int post[], int &inx_p)
    {
        if(inx_p < 0 || i0 > i1) return NULL;

        TreeNode *pRoot = new TreeNode(post[inx_p]);

        int iRoot = std::find(in, in + insize, post[inx_p--]) - in;
        TreeNode *pRight = _buildTree(in, iRoot + 1, i1, insize, post, inx_p);
        pRoot->right = pRight;
        TreeNode *pLeft= _buildTree(in, i0, iRoot-1, insize, post, inx_p);
        pRoot->left = pLeft;
        return pRoot;
    }
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        int *in = new int[inorder.size()];
        std::copy(inorder.begin(), inorder.end(), in);
        int *post = new int[postorder.size()];
        std::copy(postorder.begin(), postorder.end(), post);
        int inx = postorder.size() - 1;
        return _buildTree(in, 0, inorder.size()-1, inorder.size(), post, inx);
    }
};
posted on 2014-07-23 04:13  Tonix  阅读(144)  评论(0)    收藏  举报