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958. Check Completeness of a Binary Tree

Description

Given the root of a binary tree, determine if it is a complete binary tree.

In a complete binary tree, every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

 

Example 1:

Input: root = [1,2,3,4,5,6]
Output: true
Explanation: Every level before the last is full (ie. levels with node-values {1} and {2, 3}), and all nodes in the last level ({4, 5, 6}) are as far left as possible.

Example 2:

Input: root = [1,2,3,4,5,null,7]
Output: false
Explanation: The node with value 7 isn't as far left as possible.

 

Constraints:

• The number of nodes in the tree is in the range [1, 100].
• 1 <= Node.val <= 1000

Solutions

• Java
• C++
• Python
• Go

• /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode() {}
   *     TreeNode(int val) { this.val = val; }
   *     TreeNode(int val, TreeNode left, TreeNode right) {
   *         this.val = val;
   *         this.left = left;
   *         this.right = right;
   *     }
   * }
   */
  class Solution {
      public boolean isCompleteTree(TreeNode root) {
          Deque<TreeNode> q = new LinkedList<>();
          q.offer(root);
          while (q.peek() != null) {
              TreeNode node = q.poll();
              q.offer(node.left);
              q.offer(node.right);
          }
          while (!q.isEmpty() && q.peek() == null) {
              q.poll();
          }
          return q.isEmpty();
      }
  }
  

• /**
   * Definition for a binary tree node.
   * struct TreeNode {
   *     int val;
   *     TreeNode *left;
   *     TreeNode *right;
   *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
   *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
   *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
   * };
   */
  class Solution {
  public:
      bool isCompleteTree(TreeNode* root) {
          queue<TreeNode*> q{ {root} };
          while (q.front()) {
              root = q.front();
              q.pop();
              q.push(root->left);
              q.push(root->right);
          }
          while (!q.empty() && !q.front()) q.pop();
          return q.empty();
      }
  };
  

• # Definition for a binary tree node.
  # class TreeNode:
  #     def __init__(self, val=0, left=None, right=None):
  #         self.val = val
  #         self.left = left
  #         self.right = right
  class Solution:
      def isCompleteTree(self, root: TreeNode) -> bool:
          q = deque([root])
          while q:
              node = q.popleft()
              if node is None:
                  break
              q.append(node.left)
              q.append(node.right)
          return all(node is None for node in q)
  
  

• /**
   * Definition for a binary tree node.
   * type TreeNode struct {
   *     Val int
   *     Left *TreeNode
   *     Right *TreeNode
   * }
   */
  func isCompleteTree(root *TreeNode) bool {
  	q := []*TreeNode{root}
  	for q[0] != nil {
  		root = q[0]
  		q = q[1:]
  		q = append(q, root.Left)
  		q = append(q, root.Right)
  	}
  	for len(q) > 0 && q[0] == nil {
  		q = q[1:]
  	}
  	return len(q) == 0
  }
  

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