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04.03. List of Depth

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

Solution 1: BFS Level Order Traversal

We can use the BFS level order traversal method. For each level, we store the values of the current level's nodes into a list, and then add the list to the result array.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the number of nodes in the binary tree.

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# 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 = deque([tree])
        while q:
            dummy = cur = ListNode(0)
            for _ in range(len(q)):
                node = q.popleft()
                cur.next = ListNode(node.val)
                cur = cur.next
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(dummy.next)
        return ans

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/**
 * 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> ans = new ArrayList<>();
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(tree);
        while (!q.isEmpty()) {
            ListNode dummy = new ListNode(0);
            ListNode cur = dummy;
            for (int k = q.size(); k > 0; --k) {
                TreeNode node = q.poll();
                cur.next = new ListNode(node.val);
                cur = cur.next;
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
            ans.add(dummy.next);
        }
        return ans.toArray(new ListNode[0]);
    }
}

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/**
 * 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{{tree}};
        while (!q.empty()) {
            ListNode* dummy = new ListNode(0);
            ListNode* cur = dummy;
            for (int k = q.size(); k; --k) {
                TreeNode* node = q.front();
                q.pop();
                cur->next = new ListNode(node->val);
                cur = cur->next;
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
            ans.push_back(dummy->next);
        }
        return ans;
    }
};

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/**
 * 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) (ans []*ListNode) {
    q := []*TreeNode{tree}
    for len(q) > 0 {
        dummy := &ListNode{}
        cur := dummy
        for k := len(q); k > 0; k-- {
            node := q[0]
            q = q[1:]
            cur.Next = &ListNode{Val: node.Val}
            cur = cur.Next
            if node.Left != nil {
                q = append(q, node.Left)
            }
            if node.Right != nil {
                q = append(q, node.Right)
            }
        }
        ans = append(ans, dummy.Next)
    }
    return
}

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/**
 * 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: Array<ListNode | null> = [];
    const q: Array<TreeNode | null> = [tree];
    while (q.length) {
        const dummy = new ListNode();
        let cur = dummy;
        for (let k = q.length; k; --k) {
            const { val, left, right } = q.shift()!;
            cur.next = new ListNode(val);
            cur = cur.next;
            left && q.push(left);
            right && q.push(right);
        }
        ans.push(dummy.next);
    }
    return ans;
}

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// 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::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn list_of_depth(tree: Option<Rc<RefCell<TreeNode>>>) -> Vec<Option<Box<ListNode>>> {
        let mut res = vec![];
        if tree.is_none() {
            return res;
        }
        let mut q = VecDeque::new();
        q.push_back(tree);
        while !q.is_empty() {
            let n = q.len();
            let mut demmy = Some(Box::new(ListNode::new(0)));
            let mut cur = &mut demmy;
            for _ in 0..n {
                if let Some(node) = q.pop_front().unwrap() {
                    let mut node = node.borrow_mut();
                    if node.left.is_some() {
                        q.push_back(node.left.take());
                    }
                    if node.right.is_some() {
                        q.push_back(node.right.take());
                    }
                    cur.as_mut().unwrap().next = Some(Box::new(ListNode::new(node.val)));
                    cur = &mut cur.as_mut().unwrap().next;
                }
            }
            res.push(demmy.as_mut().unwrap().next.take());
        }
        res
    }
}

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/* class TreeNode {
*    var val: Int
*    var left: TreeNode?
*    var right: TreeNode?
*
*    init(_ val: Int) {
*        self.val = val
*        self.left = nil
*        self.right = nil
*    }
*  }
*/

/* class ListNode {
*    var val: Int
*    var next: ListNode?
*
*    init(_ val: Int) {
*        self.val = val
*        self.next = nil
*    }
*  }
*/

class Solution {
    func listOfDepth(_ tree: TreeNode?) -> [ListNode?] {
        var ans = [ListNode?]()
        guard let tree = tree else { return ans }

        var q = [TreeNode]()
        q.append(tree)

        while !q.isEmpty {
            let dummy = ListNode(0)
            var cur = dummy
            for _ in 0..<q.count {
                let node = q.removeFirst()
                cur.next = ListNode(node.val)
                cur = cur.next!

                if let left = node.left {
                    q.append(left)
                }
                if let right = node.right {
                    q.append(right)
                }
            }
            ans.append(dummy.next)
        }

        return ans
    }
}

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