1. 题目

将一个二叉树按照中序遍历转换成双向链表。

样例

样例 1：
输入:4/ \2   5/ \1   3		输出: 1<->2<->3<->4<->5样例 2：
输入:3/ \4   1输出:4<->3<->1

https://www.lintcode.com/problem/convert-binary-tree-to-doubly-linked-list/description

2. 解题

• 非递归中序遍历，使用栈

/*** Definition of Doubly-ListNode* class DoublyListNode {* public:*     int val;*     DoublyListNode *next, *prev;*     DoublyListNode(int val) {*         this->val = val;*         this->prev = this->next = NULL;*     }* } * Definition of TreeNode:* class TreeNode {* public:*     int val;*     TreeNode *left, *right;*     TreeNode(int val) {*         this->val = val;*         this->left = this->right = NULL;*     }* }*/class Solution {
public:/*** @param root: The root of tree* @return: the head of doubly list node*/DoublyListNode * bstToDoublyList(TreeNode * root) {// write your code hereif(!root) return NULL;stack<TreeNode*> stk;DoublyListNode *h = new DoublyListNode(-1), *pre = h;while(!stk.empty() || root){while(root){stk.push(root);root = root->left;}root = stk.top();stk.pop();DoublyListNode *cur = new DoublyListNode(root->val);pre->next = cur;cur->prev = pre;pre = cur;root = root->right;}DoublyListNode *head = h->next;h->next = NULL;head->prev = NULL;delete h;return head;}
};

50 ms C++

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