题解 | #二叉搜索树与双向链表#

非递归：

struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    TreeNode *head;
    TreeNode *pre;
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if(!pRootOfTree)
            return NULL;
        stack<TreeNode *>s;
        while(pRootOfTree||!s.empty()){
            while(pRootOfTree){
                s.push(pRootOfTree);
                pRootOfTree=pRootOfTree->left;
            }
            pRootOfTree=s.top();
            s.pop();
            if(!pre)
                pre=head=pRootOfTree;
            else {
                pre->right=pRootOfTree;
                pRootOfTree->left=pre;
                pre=pRootOfTree;
            }
            pRootOfTree=pRootOfTree->right;
        }
        return head;
    }
};