代码随想录训练营第三十期|第十五天|二叉树part02|层序遍历 10 ● 226.翻转二叉树 ● 101.对称二叉树 2
层序遍历 10
102. 二叉树的层序遍历 - 力扣（LeetCode）
class Solution {
public List> levelOrder(TreeNode root) {
List> res = new ArrayList<>();
if (root == null) return res;
Queue queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
List list = new ArrayList<>();
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
list.add(cur.val);
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
res.add(new ArrayList<>(list));
}
return res;
}
} 107. 二叉树的层序遍历 II - 力扣（LeetCode）
class Solution {
public List> levelOrderBottom(TreeNode root) {
List> res = new ArrayList<>();
if (root == null) return res;
Queue queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
List list = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
list.add(cur.val);
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
res.add(0,list);
}
return res;
}
} 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 rightSideView(TreeNode root) {
List res = new ArrayList<>();
if (root == null) return res;
Queue queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
if (i == size - 1) res.add(cur.val);
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
}
return res;
}
} 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 averageOfLevels(TreeNode root) {
List res = new ArrayList<>();
if (root == null) return res;
Queue queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
double sum = 0;
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
sum += cur.val;
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
res.add(sum / size);
}
return res;
}
} 429. N 叉树的层序遍历 - 力扣（LeetCode）
/*
// Definition for a Node.
class Node {
public int val;
public List children;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, List _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public List> levelOrder(Node root) {
List> res = new ArrayList<>();
if (root == null) return res;
Queue queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
List list = new ArrayList<>();
for (int i = 0; i < size; i++) {
Node cur = queue.poll();
list.add(cur.val);
if (cur.children != null) {
for (Node c : cur.children) {
queue.add(c);
}
}
}
res.add(new ArrayList<>(list));
}
return res;
}
} 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 largestValues(TreeNode root) {
List res = new ArrayList<>();
if (root == null) return res;
Queue queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
int max = Integer.MIN_VALUE;
List list = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
list.add(cur.val);
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
for (int n : list) {
max = Math.max(max, n);
}
res.add(max);
}
return res;
}
} 116. 填充每个节点的下一个右侧节点指针 - 力扣（LeetCode）
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
if (root == null) return null;
Queue queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
Node cur = queue.poll();
if (i < size - 1) {
cur.next = queue.peek();
}
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
}
return root;
}
} 117. 填充每个节点的下一个右侧节点指针 II - 力扣（LeetCode）
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
Queue queue = new LinkedList<>();
if (root == null) return null;
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
Node cur = queue.poll();
if (i < size - 1) {
cur.next = queue.peek();
}
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
}
return root;
}
} 104. 二叉树的最大深度 - 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 int maxDepth(TreeNode root) {
if (root == null) return 0;
Queue queue = new LinkedList<>();
queue.add(root);
int level = 0;
while (!queue.isEmpty()) {
int size = queue.size();
level++;
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
}
return level;
}
} 111. 二叉树的最小深度 - 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 int minDepth(TreeNode root) {
if (root == null) return 0;
Queue queue = new LinkedList<>();
int level = 0;
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
level++;
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
if (cur.left == null && cur.right == null) {
return level;
}
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
}
return level;
}
} 226. 翻转二叉树 - 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 TreeNode invertTree(TreeNode root) {
if (root == null) return root;
swap(root);
invertTree(root.left);
invertTree(root.right);
return root;
}
private void swap(TreeNode root) {
TreeNode tmp = root.left;
root.left = root.right;
root.right = tmp;
}
}101. 对称二叉树 - 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 boolean isSymmetric(TreeNode root) {
if (root == null) return true;
return dfs(root.left, root.right);
}
private boolean dfs(TreeNode left, TreeNode right) {
if (left == null && right == null) return true;
if (left == null || right == null) return false;
if (left.val != right.val) return false;
return left.val == right.val && dfs(left.left, right.right) && dfs(left.right, right.left);
}
}