958.Check-Completeness-of-a-Binary-Tree.md

958. Check Completeness of a Binary Tree

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题目地址

https://leetcode.com/problems/check-completeness-of-a-binary-tree/

题目描述

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.

代码

Approach #1 BFS

/**
 * 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) {
		List<ANode> nodes = new ArrayList();
    nodes.add(new ANode(root, 1));
    int i = 0;
    while (i < nodes.size()) {
     ANode anode = nodes.get(i++);
     if (anode.node != null) {
       nodes.add(new ANode(anode.node.left, anode.code * 2));
       nodes.add(new ANode(anode.node.right, anode.code * 2 + 1))；
     }
    }
    
    return nodes.get(i - 1).code == nodes.size();
  }
}

class ANode {
  TreeNode node;
  int code;
  ANode(TreeNode node, int code) {
    this.node = node;
    this.code = code;
  }
}

Approach #2 BFS

Peek() != null

class Solution {
  public boolean isCompleteTree(TreeNode root) {
        Queue<TreeNode> bfs = new LinkedList<TreeNode>();
        bfs.offer(root);
        while (bfs.peek() != null) {
            TreeNode node = bfs.poll();
            bfs.offer(node.left);
            bfs.offer(node.right);
        }
        while (!bfs.isEmpty() && bfs.peek() == null)
            bfs.poll();
        return bfs.isEmpty();
    }
}