Question

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.

Difficulty:Medium

Category:

Analyze

Solution

class Solution {
 public:
  bool isCompleteTree(TreeNode *root) {
    vector<vector<TreeNode *>> ans;
    int count = 0;
    traverse(root, 1, ans);
    if (ans.size() < 2) return true;
    for (int i = 0; i < ans.size() - 1; ++i) {
      // cout << pow(2,i) << endl;
      if (ans[i].size() != pow(2, i)) return false;
    }

    vector<TreeNode *> prev = ans[ans.size() - 2];
    int n = prev.size();
    count = 0;
    for (int i = 0; i < n; ++i) {
      if (prev[i]->left)
        count++;
      else
        break;
      if (prev[i]->right)
        count++;
      else
        break;
    }
    if (count != ans.back().size()) return false;
    return true;
  }

 private:
  void traverse(TreeNode *root, int level, vector<vector<TreeNode *>> &res) {
    if (!root) return;

    if (level > res.size()) res.push_back(vector<TreeNode *>());
    res[level - 1].push_back(root);
    traverse(root->left, level + 1, res);
    traverse(root->right, level + 1, res);
  }
};