代码随想录算法训练营20期|第十五天|二叉树 part02|● 层序遍历 102 ● 226.翻转二叉树 ● 101.对称二叉树
- 层序遍历 102
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102. 二叉树的层序遍历
最经典的层序遍历，用Queue来存储TreeNode，用size存储个数对每一层的node
/**
* 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> levelOrder(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 path = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
path.add(cur.val);
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
res.add(new ArrayList<>(path));
}
return res;
}
} 107. 二叉树的层序遍历 II
返回结果是倒三角，代码和102一样，只是往res里加的时候，之间加到list的最前面
/**
* 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> 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 path = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
path.add(cur.val);
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
res.add(0, path);
}
return res;
}
} 199. 二叉树的右视图
其实就是return每一层的最后一个node
/**
* 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<>();
Queue queue = new LinkedList<>();
if (root == null) return res;
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
把每一层的node值相加，再除以node的个数
/**
* 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
/*
// 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 path = new ArrayList<>();
for (int i = 0; i < size; i++) {
Node cur = queue.poll();
path.add(cur.val);
if (cur.children != null) {
List childs = cur.children;
for (Node child : childs) {
if (child != null) {
queue.add(child);
}
}
}
}
res.add(new ArrayList<>(path));
}
return res;
}
} 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 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 path = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
path.add(cur.val);
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
for (int n : path) {
max = Math.max(n, max);
}
res.add(max);
}
return res;
}
} 116. 填充每个节点的下一个右侧节点指针
/*
// 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
/*
// 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
只要有node不为空，有下一次，就继续遍历，返回层数
/**
* 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;
int level = 0;
Queue queue = new LinkedList<>();
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) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
}
return level;
}
} 111. 二叉树的最小深度
/**
* 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<>();
queue.add(root);
int depth = 0;
while (!queue.isEmpty()) {
int size = queue.size();
depth++;
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
if (cur.left == null && cur.right == null) {
return depth;
}
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
}
return depth;
}
} - 226.翻转二叉树
前序或者后序都可以，中序会有的孩子被反转两次
递归：
/**
* 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 null;
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;
}
}层序遍历：在node poll出来的时候交换
/**
* 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 null;
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();
swap(cur);
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.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
/**
* 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;
return left.val == right.val && dfs(left.left, right.right) && dfs(left.right, right.left);
}
}100. 相同的树
/**
* 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 isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null) return true;
if (p == null || q == null) return false;
if (p.val != q.val) return false;
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}572. 另一棵树的子树
把sub和tree比较,返回的时候要用||，因为有一边找到就行
/**
* 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 isSubtree(TreeNode root, TreeNode subRoot) {
if (root == null && subRoot == null) return true;
if (root == null || subRoot == null) return false;
if (isSameTree(root, subRoot)) {
return true;
}
return isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);
}
private boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null) return true;
if (p == null || q == null) return false;
if (p.val != q.val) return false;
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}