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Check Completeness of a Binary Tree⚓︎

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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 2^h 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⚓︎

BFS⚓︎

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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) {
        if (!root) return true;

        bool nullNodeFound = false;
        queue<TreeNode*> que;
        que.push(root);

        while (!que.empty()) {
            auto node = que.front();
            que.pop();

            if (!node) {
                nullNodeFound = true;
            } else {
                if (nullNodeFound) return false;
                que.push(node->left);
                que.push(node->right);
            }
        }
        return true;
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(n)\).

DFS⚓︎

Starting with the root node and assigning it an index of 0, we can use the above property to assign indices to all the other nodes in the tree. Let n represent the total number of nodes in the tree. As we saw above, the assigned index of every node must be smaller than or equal to n for the given tree to form a complete binary tree.

If the index of a node is greater or equal to n, it means a node is missing from the first n indices. In such a case, the tree is not a complete binary tree. The array representation of such a binary tree will have at least one null node in between non-null nodes.

If index < n, we proceed to its children. We use index as 2 * index + 1 for the left child node.left and 2 * index + 2 for the right child. Similarly, we determine whether or not the index of the both the children is greater than n or not.

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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 {
private:
    int countNodes(TreeNode* root) {
        if (!root) return 0;
        return 1 + countNodes(root->left) + countNodes(root->right);
    }

    bool dfs(TreeNode* node, int index, int n) {
        if (!node) return true;

        if (index >= n) return false;

        return dfs(node->left, 2 * index + 1, n) 
            && dfs(node->right, 2 * index + 2, n);
    }

public:
    bool isCompleteTree(TreeNode* root) {
        return dfs(root, 0, countNodes(root));
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(n)\).