题目描述

思路分析

dfs保存遍历结果，重新新建链表进行连接，不推荐。
但是可以根据二叉树的特点，left和right直接dfs改变指向。

代码展示

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;

    Node() {}

    Node(int _val) {
        val = _val;
        left = NULL;
        right = NULL;
    }

    Node(int _val, Node* _left, Node* _right) {
        val = _val;
        left = _left;
        right = _right;
    }
};
*/
class Solution {
    Node* prev = nullptr;
    Node* head = nullptr;
public:
    Node* treeToDoublyList(Node* root) {
        if (!root) {
            return root;
        }
        InOrder(root);
        head->left = prev;
        prev->right = head;
        return head;
    }
    void InOrder(Node* cur) {
        if (cur == nullptr) {
            return;
        }
        InOrder(cur->left);
        //改变指向
        if (prev == nullptr) {
            prev = cur;
            head = cur;//只进入一次
        }else {
            prev->right = cur;
            cur->left = prev;
            prev = cur;
        }
        InOrder(cur->right);
    }
};

运行结果

时间复杂度O(N)
空间复杂度O(N)