Skip to content

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

Solution 1

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
# 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)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
/**
 * 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();
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
/**
 * 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();
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
/**
 * 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
}

Comments