题解 | #二叉搜索树与双向链表#

//利用栈辅助遍历，时间复杂度O(n)，空间复杂度O(n)
/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
#include <cstddef>
#include <stack>
class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if (pRootOfTree==NULL) {
			return NULL;
		}
		stack<TreeNode*> s;
		TreeNode* pre;
		TreeNode* head;
		bool flag=true;
		while (pRootOfTree || !s.empty()) {
			while (pRootOfTree) {
				s.push(pRootOfTree);
				pRootOfTree=pRootOfTree->left;
			}
			pRootOfTree=s.top();
			s.pop();
			if (flag) {
				head=pRootOfTree;
				pre=head;
				flag=false;
			}else{
				pre->right=pRootOfTree;
				pRootOfTree->left=pre;
				pre=pRootOfTree;
			}
			pRootOfTree=pRootOfTree->right;	
		
		}
		return head;
		
    }
};