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:
![](https://spcdn.pages.dev/leetcode/problems/0958.Check%20Completeness%20of%20a%20Binary%20Tree/images/complete-binary-tree-1.png)
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:
![](https://spcdn.pages.dev/leetcode/problems/0958.Check%20Completeness%20of%20a%20Binary%20Tree/images/complete-binary-tree-2.png)
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#
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| # 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)
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| /**
* 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();
}
}
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| /**
* 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();
}
};
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| /**
* 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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