• 【算法leetcode】面试题 04.03. 特定深度节点链表（多语言实现）

    文章目录

    面试题 04.03. 特定深度节点链表：

    给定一棵二叉树，设计一个算法，创建含有某一深度上所有节点的链表（比如，若一棵树的深度为 D，则会创建出 D 个链表）。返回一个包含所有深度的链表的数组。

    样例 1：

    输入：
	[1,2,3,4,5,null,7,8]
	
	        1
	       /  \ 
	      2    3
	     / \    \ 
	    4   5    7
	   /
	  8

输出：
	[[1],[2,3],[4,5,7],[8]]

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    分析

    • 面对这道算法题目，二当家的陷入了沉思。
    • 如果对二叉树和链表比较熟悉，就会明白其实就是二叉树的层序遍历，每一层组合成一条链表。

    题解

    rust

    // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
// 
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
// 
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    pub fn list_of_depth(tree: Option<Rc<RefCell<TreeNode>>>) -> Vec<Option<Box<ListNode>>> {
        let mut ans: Vec<Option<Box<ListNode>>> = Vec::new();

        let mut queue = std::collections::VecDeque::new();
        queue.push_back(tree);

        while !queue.is_empty() {
            let mut head = Some(Box::new(ListNode::new(0)));
            let mut tail = head.as_mut();
            let size = queue.len();
            for _ in 0..size {
                let node = queue.pop_front().unwrap().unwrap();
                let mut node = node.borrow_mut();
                if node.left.is_some() {
                    queue.push_back(node.left.take());
                }
                if node.right.is_some() {
                    queue.push_back(node.right.take());
                }
                tail.as_mut().unwrap().next = Some(Box::new(ListNode::new(node.val)));
                tail = tail.unwrap().next.as_mut();
            }
            ans.push(head.as_mut().unwrap().next.take());
        }

        ans
    }
}

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    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 {
    var ans []*ListNode
	queue := []*TreeNode{tree}

	for len(queue) > 0 {
		head := &ListNode{}
		tail := head
		size := len(queue)
		for i := 0; i < size; i++ {
			node := queue[i]
			if node.Left != nil {
				queue = append(queue, node.Left)
			}
			if node.Right != nil {
				queue = append(queue, node.Right)
			}
			tail.Next = &ListNode{Val: node.Val}
			tail = tail.Next
		}
		ans = append(ans, head.Next)
		queue = queue[size:]
	}

	return ans
}

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    typescript

    /**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function listOfDepth(tree: TreeNode | null): Array<ListNode | null> {
    const ans = [];

	const queue = [tree];

	while (queue.length > 0) {
		const head = new ListNode();
		let tail = head;
		const size = queue.length;
		for (let i = 0; i < size; ++i) {
			const { val, left, right } = queue.shift();
			left && queue.push(left);
			right && queue.push(right);
			tail.next = new ListNode(val);
			tail = tail.next;
		}
		ans.push(head.next);
	}

	return ans;
};

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    python

    # 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]:
        ans = []
        q = collections.deque()
        q.append(tree)
        while len(q) > 0:
            head = ListNode()
            tail = head
            size = len(q)
            for _ in range(size):
                node = q.popleft()
                node.left and q.append(node.left)
                node.right and q.append(node.right)
                tail.next = ListNode(node.val)
                tail = tail.next
            ans.append(head.next)
        return ans


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    c

    /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

int getDepth(struct TreeNode* tree) {
    if (!tree) {
        return 0;
    }
    int leftDepth = getDepth(tree->left);
    int rightDepth = getDepth(tree->right);
    return fmax(leftDepth, rightDepth) + 1;
}

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
struct ListNode** listOfDepth(struct TreeNode* tree, int* returnSize){
    int depth = getDepth(tree);
    struct ListNode **ans = malloc(depth * sizeof(struct ListNode *));
    *returnSize = 0;
    struct TreeNode *queue[(int) pow(2, depth) - 1];
    queue[0] = tree;
    int start = 0;
    int end = 1;
    while (start < end) {
        struct ListNode head = {};
        struct ListNode *tail = &head;
        int curEnd = end;
        while (start < curEnd) {
            struct TreeNode *node = queue[start++];
            if (node->left) {
                queue[end++] = node->left;
            }
            if (node->right) {
                queue[end++] = node->right;
            }
            tail->next = malloc(sizeof(struct ListNode));
            tail->next->val = node->val;
            tail->next->next = NULL;
            tail = tail->next;
        }
        ans[(*returnSize)++] = head.next;
    }

    return ans;
}

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    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;

        queue<TreeNode *> q;
        q.push(tree);

        while (q.size() > 0) {
            ListNode head = ListNode(0);
            ListNode *tail = &head;
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                TreeNode *node = q.front();
                q.pop();
                if (node->left != NULL) {
                    q.push(node->left);
                }
                if (node->right != NULL) {
                    q.push(node->right);
                }
                tail->next = new ListNode(node->val);
                tail = tail->next;
            }
            ans.emplace_back(head.next);
        }

        return ans;
    }
};

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    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) {
        List<ListNode> list = new ArrayList<>();

        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(tree);

        while (!queue.isEmpty()) {
            ListNode head = new ListNode();
            ListNode tail = head;
            int      size = queue.size();
            for (int i = 0; i < size; ++i) {
                TreeNode node = queue.poll();
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
                tail.next = new ListNode(node.val);
                tail = tail.next;
            }
            list.add(head.next);
        }

        ListNode[] ans = new ListNode[list.size()];
        list.toArray(ans);

        return ans;
    }
}

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    原题传送门：https://leetcode.cn/problems/list-of-depth-lcci/submissions/

    非常感谢你阅读本文~
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    希望我们大家都能每天进步一点点~
    本文由 二当家的白帽子：https://le-yi.blog.csdn.net/ 博客原创~

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  • 原文地址：https://blog.csdn.net/leyi520/article/details/125407346