lt. 102. 二叉树的层序遍历
[案例需求]
[思路分析]
- 二叉树的层序遍历可真是二叉树题目中的大明星了哈哈. 彻底的理解这道题有助于后面多达十几道不同类型题目的求解;
- 具体分析: 点我
[代码实现]
/**
* 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<List<Integer>> levelOrder(TreeNode root) {
//BFS
List<List<Integer>> lists = new ArrayList<>();
Queue<TreeNode> queue = new ArrayDeque<>();
if(root != null) queue.offer(root);
while(!queue.isEmpty()){
//
int n = queue.size(); // 注意queue的size是不断变化的, 所以不适合把这个queue.size写在for循环中;
List<Integer> list = new ArrayList<>();
for(int i = 0; i < n; i++){
root = queue.poll();
list.add(root.val);
if(root.left != null){
queue.offer(root.left);
}
if(root.right != null){
queue.offer(root.right);
}
}
lists.add(list);
}
return lists;
}
}
lt.103. 二叉树的锯齿形层序遍历
[案例需求]
[思路分析一, 学弱解法(完全借鉴层序遍历)]
- 本题跟层序遍历解法其实基本上是相同的, 都是把当前层的结点先存入队列, 然后再从队列中取出, 一边取出, 一边把正在取出的结点的左右孩子结点存进去, 而且, 为了达到每一轮都遍历一层的效果, 在取出过程开始之前, 我们先算出该层在queue中的size大小, 然后循环size次进行取出;
- 为了达到本体的Z字形遍历的效果, 最简单的方法就是加入一个奇偶判断, 当为奇数时, 从左到右取出元素, 当为偶数时, 从右到左取出元素, 如何实现的呢? 就是在把从queue中取出的结点倒序添加到list中即可 (list.add(0, 结点值))
[代码实现]
/**
* 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<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> lists = new ArrayList<>();
Queue<TreeNode> queue = new ArrayDeque<>();
if(root != null) queue.add(root);
int levelNumber = 0;
//
while(!queue.isEmpty()){
List<Integer> level = new ArrayList<>();
int size = queue.size();
if(levelNumber % 2 == 0){
for(int i = 0; i < size; i++){
TreeNode node = queue.poll();
level.add(node.val);
if(node.left != null){
queue.add(node.left);
}
if(node.right != null){
queue.add(node.right);
}
}
}else{
for(int i = 0; i < size; i++){
TreeNode node = queue.poll();
level.add(0, node.val); //遍历的结点反向添加到list中, 就变为了倒序遍历本层
if(node.left != null){
queue.add(node.left);
}
if(node.right != null){
queue.add(node.right);
}
}
}
++levelNumber;
lists.add(level);
}
return lists;
}
}
/**
* 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<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> lists = new ArrayList<>();
Queue<TreeNode> queue = new ArrayDeque<>();
if(root != null) queue.add(root);
int levelNumber = 0;
boolean reverse = false;
//
while(!queue.isEmpty()){
List<Integer> level = new ArrayList<>();
int size = queue.size();
for(int i = 0; i < size; i++){
TreeNode node = queue.poll();
level.add(node.val);
if(node.left != null){
queue.add(node.left);
}
if(node.right != null){
queue.add(node.right);
}
}
if(reverse){Collections.reverse(level);}
reverse = !reverse;
lists.add(level);
}
return lists;
}
}
[思路分析二, 使用双端队列]
- 从左到右的, 双端队列添加结点从头部添加
- 从右到左的, 结点从尾部添加, 应该不难理解哈
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> ans = new LinkedList<List<Integer>>();
if (root == null) {
return ans;
}
Queue<TreeNode> nodeQueue = new ArrayDeque<TreeNode>();
nodeQueue.offer(root);
boolean isOrderLeft = true;
while (!nodeQueue.isEmpty()) {
Deque<Integer> levelList = new LinkedList<Integer>();
int size = nodeQueue.size();
for (int i = 0; i < size; ++i) {
TreeNode curNode = nodeQueue.poll();
if (isOrderLeft) {
levelList.offerLast(curNode.val);
} else {
levelList.offerFirst(curNode.val);
}
if (curNode.left != null) {
nodeQueue.offer(curNode.left);
}
if (curNode.right != null) {
nodeQueue.offer(curNode.right);
}
}
ans.add(new LinkedList<Integer>(levelList));
isOrderLeft = !isOrderLeft;
}
return ans;
}
}
[代码实现]
lt.107. 二叉树的层序遍历 II
[案例需求]
[思路分析]
- 102 题, 缝缝补补还能用哈哈哈
[代码实现]
/**
* 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<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> lists = new ArrayList<>();
if(root == null)return lists;
Queue<TreeNode> queue = new ArrayDeque<>();
queue.add(root);
while(!queue.isEmpty()){
int size = queue.size();
List<Integer> list = new ArrayList<>();
for(int i = 0; i < size; i++){
root = queue.poll();
list.add(root.val);
if(root.left != null) queue.add(root.left);
if(root.right != null) queue.add(root.right);
}
lists.add(0, list);
}
// 上面写成 lists.add(list), 这里加上下面这一句也是可行的
// Collections.reverse(lists);
return lists;
}
}
lt.199. 二叉树的右视图
[案例需求]
[思路分析]
- 这题更有意思, 但也不过是层序遍历换汤不换药罢了哈哈哈;
在层序遍历中, 我们一边把每一层的结点取出来放入到list中, 一边还要把这个节点下一层的左右孩子结点存入到队列中, 注意这个过程是循环的, 循环的次数就是在取结点之前queue的size数(他也是每一层结点的个数);
- 题目要求我们只打印出每层的最右边的结点, 那我们直接只把每一层最右边的结点存入list就是了, 也就是当循环到最后一个结点时, 才把这个节点添加到list中去
(if(i == size - 1) list.add(结点))
[代码实现]
/**
* 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> rightSideView(TreeNode root) {
//右视图, 每次只记录每一层最右边的数, 是不是就是
//层寻遍历中, 在添加结点到list时, 我们只需要添加最后一个出队列的元素即可
List<Integer> list = new ArrayList<>();
if(root == null)return list;
Queue<TreeNode> queue = new ArrayDeque<>();
queue.add(root);
while(!queue.isEmpty()){
int size = queue.size();
for(int i = 0; i < size; i++){
root = queue.poll();
if(i == size - 1) list.add(root.val);
if(root.left != null)queue.add(root.left);
if(root.right != null)queue.add(root.right);
}
}
return list;
}
}
lt.637. 二叉树的层平均值
[案例需求]
[思路分析]
- 朴实无华的BFS, 层序遍历
- 注意数据类型范围溢出问题!
[代码实现一, BFS]
/**
* 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> list = new ArrayList<>();
if(root == null)return list;
Queue<TreeNode> queue = new ArrayDeque<>();
queue.add(root);
while(!queue.isEmpty()){
int size = queue.size();
double sum = 0;
for(int i = 0; i < size; i++){
root = queue.poll();
sum += root.val;
if(root.left != null)queue.add(root.left);
if(root.right != null)queue.add(root.right);
}
double avrg = sum / size;
list.add(avrg);
}
return list;
}
}
lt.429, N 叉树的层序遍历
[案例需求]
[思路分析]
- 一招鲜哎我吃遍天;
- 结合Node的内部类声明, 我们可以发现所谓的N叉树其实就是把多个节点存储到一个List 中去的, 大体上跟层序遍历完全一致, 只是在原来遍历root 的左右结点基础上, 变为了遍历List , 添加到队列中去的.
[代码实现]
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> children;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, List<Node> _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public List<List<Integer>> levelOrder(Node root) {
List<List<Integer>> lists = new ArrayList<>();
if(root == null)return lists;
Queue<Node> queue = new ArrayDeque<>();
queue.add(root);
while(!queue.isEmpty()){
int size = queue.size();
List<Integer> list = new ArrayList<>();
for(int i = 0; i < size; i++){
root = queue.poll();
list.add(root.val);
if(root.children != null){
for(Node node : root.children){
queue.add(node);
}
}
}
lists.add(list);
}
return lists;
}
}
X.<tag-数组和二分查找>-lt.xx-xxxxxx + lt.xx-xxxxxx
lt.515. 在每个树行中找最大值
[案例需求]
[思路分析]
- 题目要求是记录每一层的最大值, 我们只需要在层序遍历的基础上, 比较较大的值, 待该层遍历完成后,
- 再把每层的最大值存入到list中即可.
[代码实现]
/**
* 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> list = new ArrayList<>();
if(root == null) return list;
Queue<TreeNode> queue = new ArrayDeque<>();
queue.add(root);
while(!queue.isEmpty()){
int size = queue.size();
int max = Integer.MIN_VALUE;
for(int i = 0; i < size; i++){
//每一层结点
root = queue.poll();
max = Math.max(root.val, max);
if(root.left != null)queue.add(root.left);
if(root.right != null)queue.add(root.right);
}
list.add(max);
}
return list;
}
}
lt.240-搜索二维矩阵 ||
[案例需求]
[思路分析]
[代码实现]
lt.240-搜索二维矩阵 ||
[案例需求]
[思路分析]
[代码实现]