[剑指Offer] 26.二叉搜索树与双向链表

【思路】因为二叉搜索树的中序遍历就是递增排列的，所以只要在中序遍历时将每个结点放入vector中，再分别为每个结点的左右指针赋值即可。

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution
{
public:
    vector<TreeNode *> nodeVector;

    void inOrder(TreeNode* root)
    {
        if(root->left)
            inOrder(root->left);
        nodeVector.push_back(root);
        if(root->right)
            inOrder(root->right);
    }

    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if((pRootOfTree == NULL) || (!pRootOfTree->left) && (!pRootOfTree->right))
            return pRootOfTree;
        inOrder(pRootOfTree);
        nodeVector[0]->left = NULL;
        nodeVector[0]->right = nodeVector[1];
        int i = 1;
        for( ; i < nodeVector.size() - 1; i ++)
        {
            nodeVector[i]->left = nodeVector[i - 1];
            nodeVector[i]->right = nodeVector[i + 1];
        }
        nodeVector[i]->left = nodeVector[i - 1];
        nodeVector[i]->right = NULL;
        return nodeVector[0];
    }
};