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 2h 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:

  1. The tree will have between 1 and 100 nodes.

Companies:
Facebook

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;
    }
};

Java

  • /**
     * 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;
        }
    }
    
  • // 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(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
    

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