Problem
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
Solution (Java)
/**
* 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) {
/* 1. Tree no exist? complete! */
if (root == null) {
return true;
}
/* 2. Create a queue for purposeful sequence of inserting children nodes
// (Left to right, level by level) */
LinkedList<TreeNode> queue = new LinkedList<>();
queue.add(root.left);
queue.add(root.right);
/* 3. Essential for our loop. Checks if we seen a null. */
boolean seenNull = false;
/* 4. While the queue isn't empty.
// Catch that null, and if there is a # after, it's not complete */
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
if (node == null) {
seenNull = true;
} else {
if (seenNull) {
return false;
}
queue.add(node.left);
queue.add(node.right);
}
}
/*5. If we don't catch null with # after: it's complete */
return true;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).