力扣labuladong——一刷day64

提示：文章写完后，目录可以自动生成，如何生成可参考右边的帮助文档

---

前言

---

二叉树大部分题目都可以用递归的算法解决，但少部分题目用递归比较麻烦的话，我们可以考虑使用层序遍历的方式解决。

一、力扣515. 在每个树行中找最大值

/**
 * 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 {
    public List<Integer> largestValues(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        Deque<TreeNode> deq = new ArrayDeque<>();
        deq.offerLast(root);
        while(!deq.isEmpty()){
            int size = deq.size();
            int max = Integer.MIN_VALUE;
            for(int i = 0; i < size; i ++){
                TreeNode cur = deq.pollFirst();
                if(cur.left != null){
                    deq.offerLast(cur.left);
                }
                if(cur.right != null){
                    deq.offerLast(cur.right);
                }
                max = Math.max(max, cur.val);
            }
            res.add(max);
        }
        return res;
    }
}

二、力扣637. 二叉树的层平均值

/**
 * 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 {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        Deque<TreeNode> deq = new ArrayDeque<>();
        deq.offerLast(root);
        while(!deq.isEmpty()){
            int size = deq.size();
            double sum = 0;
            for(int i = 0; i < size; i ++){
                TreeNode cur = deq.pollFirst();
                if(cur.left != null){
                    deq.offerLast(cur.left);
                }
                if(cur.right != null){
                    deq.offerLast(cur.right);
                }
                sum += cur.val;
            }
            res.add(sum / size);
        }
        return res;
    }
}

三、力扣958. 二叉树的完全性检验

/**
 * 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 {
    class Pair{
        TreeNode node;
        int id;
        public Pair(TreeNode node,int id){
            this.node = node;
            this.id = id;
        }
    }
    public boolean isCompleteTree(TreeNode root) {
        if(root == null){
            return true;
        }
        Deque<Pair> deq = new ArrayDeque<>();
        deq.offerLast(new Pair(root,1));
        int flag = 1;
        while(!deq.isEmpty()){
            int size = deq.size();
            for(int i = 0; i < size; i ++){
                Pair cur = deq.pollFirst();
                TreeNode curNode = cur.node;
                int curId = cur.id;
                if(curId != flag){
                    return false;
                }
                if(curNode.left != null){
                    deq.offerLast(new Pair(curNode.left,curId*2));
                }
                if(curNode.right != null){
                    deq.offerLast(new Pair(curNode.right,curId*2+1));
                }
                flag ++;
            }
        }
        return true;
    }
}