题目描述

• 考察对完全二叉树性质的理解
  

思路 && 代码

层级遍历法

• 一开始试了自底向上的DFS，过了90%的用例，还是考虑得不够全 = =
• 层级遍历：代码比较多，但是总体效率和方法2差不多

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {Deque<TreeNode> queue = new ArrayDeque<>();public boolean isCompleteTree(TreeNode root) {queue.offer(root);return bfs(1, 1);}public boolean bfs(int nodeNums, int except) {if(queue.size() == 0) {return true;}int counts = 0;boolean flag = false;for(int i = 0; i < nodeNums; i++) {TreeNode tempNode = queue.pop();if(tempNode.left != null) {if(flag) {return false;}queue.offer(tempNode.left);counts++;}else {flag = true;}if(tempNode.right != null) {if(flag) {return false;}queue.offer(tempNode.right);counts++;}else {flag = true;}}if(nodeNums != except && !queue.isEmpty()) {return false;}return bfs(counts, 2 * except);}
}

BFS

• 注意：要使用能存储 null 值的集合类（比如 ArrayDeque 就不行）
• 核心思路：当前节点为空时，如果后面（包括右边、下面）有节点则不是完全二叉树。

class Solution {// 完全二叉树判断：当前节点为空时，后面（包括右边、下面）不能有节点public boolean isCompleteTree(TreeNode root) {LinkedList<TreeNode> queue = new LinkedList<>();queue.offer(root);for(TreeNode cur = queue.poll(); cur != null; cur = queue.poll()) {queue.offer(cur.left);queue.offer(cur.right);}// 如果之后还有非空元素，说明不是完全二叉树while(!queue.isEmpty()) {if(queue.poll() != null) {return false;}}return true;}
}

二刷

class Solution {public boolean isCompleteTree(TreeNode root) {// jojo的奇妙做法。。不知道怎么想的，反正 A 了，问题不大！List<TreeNode> list = new LinkedList<>();Queue<TreeNode> queue = new ArrayDeque<>();queue.offer(root);list.add(root);while(!queue.isEmpty()) {TreeNode temp = queue.poll();list.add(temp.left);list.add(temp.right);if(temp.left != null) {queue.offer(temp.left);}if(temp.right != null) {queue.offer(temp.right);}}boolean flag = false;for(TreeNode temp : list) {if(temp == null) {flag = true;} else if (flag) {return false;}}return true;}
}

• 简单写法，八行代码光速解决～

class Solution {public boolean isCompleteTree(TreeNode root) {LinkedList<TreeNode> queue = new LinkedList<>();for(TreeNode cur = root; cur != null; cur = queue.poll()) {queue.offer(cur.left);queue.offer(cur.right);}while(!queue.isEmpty()) {if(queue.poll() != null) {return false;}}return true;}
}