技術標籤：LintCode及其他OJ

文章目錄

• • 1. 題目
  • 2. 解題

1. 題目

將一個二叉樹按照中序遍歷轉換成雙向連結串列。

樣例

樣例 1：
輸入:
	    4
	   / \
	  2   5
	 / \
	1   3		

輸出: 1<->2<->3<->4<->5

樣例 2：
輸入:
	    3
	   / \
	  4   1

輸出:4<->3<->1

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

2. 解題

• 非遞迴中序遍歷，使用棧

/**
  * 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 here
        if(!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++

我的CSDN部落格地址 https://michael.blog.csdn.net/

長按或掃碼關注我的公眾號（Michael阿明），一起加油、一起學習進步！