题解 | 二叉搜索树与双向链表

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
private:
	queue<TreeNode*> q;
	void midTrace(TreeNode* root){
		if(root == nullptr) return;
		midTrace(root->left);
		q.push(root);
		midTrace(root->right);
		return;
	}
public:
    TreeNode* Convert(TreeNode* pRootOfTree) {
        midTrace(pRootOfTree);
		TreeNode* head = nullptr;
		TreeNode* pre = nullptr;
		while(!q.empty()){
			TreeNode* tmp = q.front();
			q.pop();
			if(pre != nullptr){
				tmp->left=pre;
				pre->right= tmp;
				pre = tmp;
			}
			else{
				pre =tmp;
				head = tmp;
			}
		}
		return head;
    }
};