Formatted question description: https://leetcode.ca/all/958.html

958. Check Completeness of a Binary Tree (Medium)

Given a binary tree, determine if it is a complete binary tree.

Definition of a complete binary tree from Wikipedia:
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 2^h nodes inclusive at the last level h.

Example 1:

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

Note:

The tree will have between 1 and 100 nodes.

Companies:
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Related Topics:
Tree

Solution 1.

Use level order traversal. After visiting a node that has at least one null child, all the subsequent nodes must be leaf nodes.

// OJ: https://leetcode.com/problems/check-completeness-of-a-binary-tree/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    bool isCompleteTree(TreeNode* root) {
        if (!root) return true;
        queue<TreeNode*> q;
        q.push(root);
        bool onlyLeafNow = false;
        while (q.size()) {
            int cnt = q.size();
            while (cnt--) {
                root = q.front();
                q.pop();
                if (!root->left && root->right) return false;
                if (onlyLeafNow && (root->left || root->right)) return false;
                if (root->left) q.push(root->left);
                if (root->right) q.push(root->right);
                if (!root->left || !root->right) onlyLeafNow = true;
            }
        }
        return true;
    }
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isCompleteTree(TreeNode root) {
        if (root == null)
            return true;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        int levelNodes = 1;
        while (!queue.isEmpty()) {
            int size = queue.size();
            if (size == levelNodes) {
                boolean flag = true;
                for (int i = 0; i < size; i++) {
                    TreeNode node = queue.poll();
                    TreeNode left = node.left, right = node.right;
                    if (left == null && right != null)
                        return false;
                    if (left != null)
                        queue.offer(left);
                    if (right != null)
                        queue.offer(right);
                    if (!flag && (left != null || right != null))
                        return false;
                    if (flag && (left == null || right == null))
                        flag = false;
                }
            } else {
                for (int i = 0; i < size; i++) {
                    TreeNode node = queue.poll();
                    if (node.left != null || node.right != null)
                        return false;
                }
            }
            levelNodes <<= 1;
        }
        return true;
    }
}

############

/**
 * 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();
    }
}

// OJ: https://leetcode.com/problems/check-completeness-of-a-binary-tree/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    bool isCompleteTree(TreeNode* root) {
        if (!root) return true;
        queue<TreeNode*> q;
        q.push(root);
        bool onlyLeafNow = false;
        while (q.size()) {
            int cnt = q.size();
            while (cnt--) {
                root = q.front();
                q.pop();
                if (!root->left && root->right) return false;
                if (onlyLeafNow && (root->left || root->right)) return false;
                if (root->left) q.push(root->left);
                if (root->right) q.push(root->right);
                if (!root->left || !root->right) onlyLeafNow = true;
            }
        }
        return true;
    }
};

# 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.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isCompleteTree(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root: return True
        res = []
        que = collections.deque()
        que.append(root)
        hasNone = False
        while que:
            size = len(que)
            for i in range(size):
                node = que.popleft()
                if not node:
                    hasNone = True
                    continue
                if hasNone:
                    return False
                que.append(node.left)
                que.append(node.right)
        return True

/**
 * 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
}

958 - Check Completeness of a Binary Tree

958. Check Completeness of a Binary Tree (Medium)

Solution 1.

All Problems

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958. Check Completeness of a Binary Tree (Medium)

Solution 1.

All Problems

All Solutions