Q 4.4 creates a linked list of all the nodes at each depth_create a linked list from file fp in which each no-优快云博客

本文链接：https://blog.youkuaiyun.com/xinxinxv/article/details/42928727

本文介绍了一种将二叉搜索树转换为多个层级链表的算法。通过两种方法实现：一种是递归地逐层建立链表；另一种是使用广度优先搜索（BFS）的方式迭代实现。这两种方法都能有效地处理任意深度的二叉搜索树，并创建相应数量的链表。

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Q: Given a binary search tree, design an algorithm which creates a linked list of all the nodes at each depth (i.e., if you have a tree with depth D, you’ll have D linked lists).

思路1：逐层建立链表。

遍历level层的节点，来建立level+1层的链表。即prev为已level+1链表的尾节点，则level层的节点cur->left存在的话，将其插入到链表中，即prev->next = cur->left，同样的道理查看cur->right，通过cur = cur->next,继续建立链表。依此逐层进行。

思路2： BFS。同思路1，只是把用迭代实现。

#include <iostream>
#include <vector>
using namespace std;

struct TreeLinkNode {
	int val;
	TreeLinkNode *left, *right, *next;
	TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
};

vector<TreeLinkNode*> res;

void insert(TreeLinkNode* root, int x){
    root = new TreeLinkNode(x);
    return ;
}

void connect1(TreeLinkNode *root) {
    if (!root) {
        return ;
    }
    TreeLinkNode dummy(-1);
    res.push_back(root);
    for (TreeLinkNode *cur = root, *prev = &dummy; cur; cur = cur->next) {
        if (cur->left) {
            prev->next = cur->left;
            prev = cur->left;
        }
        if (cur->right) {
            prev->next = cur->right;
            prev = cur->right;
        }
    }
    connect1(dummy.next);
}

void connect2(TreeLinkNode *root) {
    if (!root) {
        return ;
    }
    TreeLinkNode *cur = root;
    while (cur) {
        TreeLinkNode *next = NULL;
        TreeLinkNode *prev = NULL;
        while (cur) {
            if (!next) {
                if (cur->left) {
                    next = cur->left;
                } else if (cur->right) {
                    next = cur->right;
                }
            }
            if (cur->left) {
                if (!prev) {
                    prev = cur->left;
                }
                prev->next = cur->left;
                prev = prev->next;
            }
            if (cur->right) {
                if (!prev) {
                    prev = cur->right;
                }
                prev->next = cur->right;
                prev = prev->next;
            }
            cur = cur->next;
        }
        cur = next;
    }
}

int main() {
	int a[7] = {1,2,3,4,5,6,7};
	TreeLinkNode *root = new TreeLinkNode(a[0]);
//	cout<<root->val<<endl;
	int i = 1;
	root->left = new TreeLinkNode(a[i++]);
	root->right = new TreeLinkNode(a[i++]);
	root->left->left = new TreeLinkNode(a[i++]);
	root->left->right = new TreeLinkNode(a[i++]);
	root->right->left = new TreeLinkNode(a[i++]);
	root->right->right = new TreeLinkNode(a[i++]);
//	cout<<root->left->val<<endl;
	connect1(root);
	for (int i = 0; i < res.size(); i++) {
		TreeLinkNode *cur = res[i];
		while (cur) {
			cout<<cur->val<<" ";
			cur = cur->next;
		}
		cout<<endl;
	}
	return 0;
}