Question

04.03. List of Depth

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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

# 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

/**
 * 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]);
  }
}

/**
 * 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;
  }
};

/**
 * 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
}

/**
 * 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;
}

// 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;
use std::collections::VecDeque;
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
  }
}