102. 二叉树的层序遍历 - 力扣（LeetCode）
使用队列进行层序遍历。
/*** 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) {List<List<Integer>>ans = new ArrayList<>();Queue<TreeNode>q = new LinkedList<>();if(root == null){return ans;}q.offer(root);int num = 1;while(!q.isEmpty()){List<Integer>temp = new ArrayList<>();num = q.size();while(num != 0){TreeNode t = q.poll();temp.add(t.val);if(t.left != null){q.offer(t.left);}if(t.right != null){q.offer(t.right);}num--;}ans.add(temp);}return ans;}
}107. 二叉树的层序遍历 II - 力扣（LeetCode）
在上一题的基础上，翻转一下ans
/*** 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>>ans = new ArrayList<>();Queue<TreeNode>q = new LinkedList<>();if(root == null){return ans;}q.offer(root);int num = 1;while(!q.isEmpty()){List<Integer>temp = new ArrayList<>();num = q.size();while(num != 0){TreeNode t = q.poll();temp.add(t.val);if(t.left != null){q.offer(t.left);}if(t.right != null){q.offer(t.right);}num--;}ans.add(temp);}Collections.reverse(ans);return ans;}
}199. 二叉树的右视图 - 力扣（LeetCode）
/*** 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) {List<Integer>ans = new ArrayList<>();Queue<TreeNode>q = new LinkedList<>();if(root == null){return ans;}q.offer(root);int num = 1;while(!q.isEmpty()){num = q.size();while((num - 1) != 0){TreeNode temp = q.poll();if(temp.left != null){q.offer(temp.left);}if(temp.right != null){q.offer(temp.right);}num--;}TreeNode t = q.poll();if(t.left != null){q.offer(t.left);}if(t.right != null){q.offer(t.right);}ans.add(t.val);}return ans;}
}637. 二叉树的层平均值 - 力扣（LeetCode）
/*** 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>ans = new ArrayList<>();Queue<TreeNode>q = new LinkedList<>();if(root == null){return ans;}q.offer(root);int num = 1;while(!q.isEmpty()){num = q.size();int tnum = num;long ad = 0;while(num != 0){TreeNode temp = q.poll();if(temp.left != null){q.offer(temp.left);}if(temp.right != null){q.offer(temp.right);}ad += temp.val;num--;}double cur = (double)ad / (double)tnum;ans.add(cur);}return ans;}
}429. N 叉树的层序遍历 - 力扣（LeetCode）
/*
// 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>>ans = new ArrayList<>();if(root == null){return ans;}Queue<Node>q = new LinkedList<>();q.offer(root);while(!q.isEmpty()){List<Integer>temp = new ArrayList<>();int num = q.size();while(num != 0){Node t = q.poll();temp.add(t.val);List<Node>list = t.children;for(Node n : list){q.offer(n);}num--;}ans.add(temp);}return ans;}
}515. 在每个树行中找最大值 - 力扣（LeetCode）
/*** 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>ans = new ArrayList<>();if(root == null){return ans;}Queue<TreeNode>q = new LinkedList<>();q.offer(root);while(!q.isEmpty()){int num = q.size();int max = Integer.MIN_VALUE;while(num != 0){TreeNode temp = q.poll();if(temp.left != null){q.offer(temp.left);}if(temp.right != null){q.offer(temp.right);}max = max > temp.val ? max : temp.val;num--;}ans.add(max);}return ans;}
}
