文章目录
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;
}
}