README_EN.md

04.03. List of Depth

中文文档

Description

Given a binary tree, design an algorithm which creates a linked list of all the nodes at each depth (e.g., if you have a tree with depth D, you'll have D linked lists). Return a array containing all the linked lists.

 

Example:

Input: [1,2,3,4,5,null,7,8]



        1

       /  \

      2    3

     / \    \

    4   5    7

   /

  8



Output: [[1],[2,3],[4,5,7],[8]]

Solutions

Level order traversal.

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def listOfDepth(self, tree: TreeNode) -> List[ListNode]:
        q = [tree]
        ans = []
        while q:
            n = len(q)
            head = ListNode(-1)
            tail = head
            for i in range(n):
                front = q.pop(0)
                node = ListNode(front.val)
                tail.next = node
                tail = node
                if front.left:
                    q.append(front.left)
                if front.right:
                    q.append(front.right)
            ans.append(head.next)
        return ans

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode[] listOfDepth(TreeNode tree) {
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(tree);
        List<ListNode> ans = new ArrayList<>();
        while (!queue.isEmpty()) {
            int n = queue.size();
            ListNode head = new ListNode(-1);
            ListNode tail = head;
            for (int i = 0; i < n; i++) {
                TreeNode front = queue.poll();
                ListNode node = new ListNode(front.val);
                tail.next = node;
                tail = node;
                if (front.left != null) {
                    queue.offer(front.left);
                }
                if (front.right != null) {
                    queue.offer(front.right);
                }
            }
            ans.add(head.next);
        }
        return ans.toArray(new ListNode[0]);
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    vector<ListNode*> listOfDepth(TreeNode* tree) {
        vector<ListNode*> ans;
        if (tree == nullptr) {
            return ans;
        }
        queue<TreeNode*> q;
        q.push(tree);
        while (!q.empty()) {
            int n = q.size();
            ListNode* head = new ListNode(-1);
            ListNode* tail = head;
            for (int i = 0; i < n; ++i) {
                TreeNode* front = q.front();
                q.pop();
                ListNode* node = new ListNode(front->val);
                tail->next = node;
                tail = node;
                if (front->left != nullptr) {
                    q.push(front->left);
                }
                if (front->right != nullptr) {
                    q.push(front->right);
                }
            }
            ans.push_back(head->next);
        }
        return ans;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func listOfDepth(tree *TreeNode) []*ListNode {
	queue := make([]*TreeNode, 0)
	queue = append(queue, tree)
	ans := make([]*ListNode, 0)
	for len(queue) > 0 {
		n := len(queue)
		head := new(ListNode)
		tail := head
		for i := 0; i < n; i++ {
			front := queue[0]
			queue = queue[1:]
			node := &ListNode{Val: front.Val}
			tail.Next = node
			tail = node
			if front.Left != nil {
				queue = append(queue, front.Left)
			}
			if front.Right != nil {
				queue = append(queue, front.Right)
			}
		}
		ans = append(ans, head.Next)
	}
	return ans
}

...