Check Completeness of a Binary Tree

update Dec 25, 16:19

LeetCode

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:

The tree will have between 1 and 100 nodes.

Basic Idea:

我们注意到，如果一个树是完全二叉树，那么在BFS的过程中，一旦我们遇到null，后面就不可以有不是null的node，我们可以利用这一性质。简单来说就是写一个BFS，用一个boolean标记是否已经见过null，如果见过，当我们再遇到非null时直接返回false；

• Java Code:

  class Solution {
    public boolean isCompleteTree(TreeNode root) {
        Deque<TreeNode> queue = new LinkedList<>();
        queue.offerLast(root);
        boolean seenNull = false;
        while (! queue.isEmpty()) {
            TreeNode currNode = queue.pollFirst();
            if (currNode == null) {
                seenNull = true;
                continue;
            } else if (seenNull) {
                return false;
            } 
            queue.offerLast(currNode.left);
            queue.offerLast(currNode.right);
        }
        return true;
    }
  }
  

还有一种dfs的解法，没有这么直观，需要仔细分情况判断不符合条件的情况。

• Java Code:

  class Solution {
    private class Ret {
        boolean isFull;
        int height;
        public Ret(int height, boolean isFull) {
            this.isFull = isFull;
            this.height = height;
        }
    }
  
    public boolean isCompleteTree(TreeNode root) {
        Ret ret = helper(root);
        if (ret == null) return false;
        else return true;
    }
  
    private Ret helper(TreeNode root) {
        if (root == null) return new Ret(0, true);
        Ret l = helper(root.left);
        Ret r = helper(root.right);
        if (l == null || r == null) return null;
        if (l.height < r.height) return null;
        if (! l.isFull && l.height == r.height) return null;
        if (! l.isFull && ! r.isFull) return null;
        if (l.height > r.height + 1) return null;
        return new Ret(l.height + 1, l.isFull && r.isFull && l.height == r.height);
    }
  }