目录
[N 叉树的层序遍历](#N 叉树的层序遍历)
宽搜（BFS）
N 叉树的层序遍历
java
/*
// 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>> ret = new ArrayList<>();
if (root == null)
return ret;
Queue<Node> q = new LinkedList<>();// 使用队列
q.add(root);
while (!q.isEmpty()) {
int sz = q.size();
List<Integer> tmp = new ArrayList<>();// 统计本层的节点信息
for (int i = 0; i < sz; i++) {
Node t = q.poll();
tmp.add(t.val);
for (Node child : t.children) {
if (child != null) {
q.add(child);// 孩子入队
}
}
}
ret.add(tmp);
}
return ret;
}
}二叉树的锯齿形层序遍历
103. 二叉树的锯齿形层序遍历 - 力扣（LeetCode）
解法：层序遍历
增加一个标记位，让偶数行 的信息逆序即可 Collections.reverse();
java
/**
* 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>> ret = new ArrayList<>();// 用来返回结果
if (root == null)
return ret;// 判空
Queue<TreeNode> q = new LinkedList<>();
q.add(root);// 用来获取二叉树中元素
int level = 1;// 标记是否是偶数行
while (!q.isEmpty()) {
int sz = q.size();
List<Integer> tmp = new LinkedList<>();// 用来记录每一行结果
for (int i = 0; i < sz; i++) {
TreeNode t = q.poll();
tmp.add(t.val);
if (t.left != null)
q.add(t.left);
if (t.right != null)
q.add(t.right);
}
// 判断是否逆序
if (level % 2 == 0)
Collections.reverse(tmp);
ret.add(tmp);
level++;
}
return ret;
}
}二叉树最大宽度
解法一：硬补 -> 超时、内存可能不够
解法二：利用数组存储二叉树的方式（堆），给节点编号
细节：①用数组来实现队列，这样队头队尾比较好找
②下标可能溢出。相减之后，即使溢出，结果也是正确的（数据存储是环形的，距离不会溢出）
java
/**
* 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 int widthOfBinaryTree(TreeNode root) {
List<Pair<TreeNode, Integer>> q = new ArrayList<>();// 用数组模拟队列
q.add(new Pair<TreeNode, Integer>(root, 1));
int ret = 0;// 记录最终结果
while (!q.isEmpty()) {
Pair<TreeNode, Integer> t1 = q.get(0);// 得到队头
Pair<TreeNode, Integer> t2 = q.get(q.size() - 1);// 得到队尾
ret = Math.max(t2.getValue() - t1.getValue() + 1, ret);
// 下一层进队
List<Pair<TreeNode, Integer>> tmp = new ArrayList<>();
for (Pair<TreeNode, Integer> t : q) {
TreeNode node = t.getKey();
int index = t.getValue();
if (node.left != null) {
tmp.add(new Pair<TreeNode, Integer>(node.left, index * 2));
}
if (node.right != null) {
tmp.add(new Pair<TreeNode, Integer>(node.right, index * 2 + 1));
}
}
q = tmp;
}
return ret;
}
}在每个树行中找最大值
515. 在每个树行中找最大值 - 力扣（LeetCode）
解法：利用层序遍历，统计出每一层最大值
java
/**
* 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> ret = new ArrayList<>();
if (root == null)
return ret;
Queue<TreeNode> q = new LinkedList<>();
q.add(root);
while (!q.isEmpty()) {
int sz = q.size();
int tmp = Integer.MIN_VALUE;
for (int i = 0; i < sz; i++) {
TreeNode t = q.poll();
tmp = Math.max(tmp, t.val);
if (t.left != null)
q.add(t.left);
if (t.right != null)
q.add(t.right);
}
ret.add(tmp);
}
return ret;
}
}