目录
102. 二叉树的层序遍历
226. 翻转二叉树
101. 对称二叉树
100. 相同的树
100是101的衍生题目。572也为101的衍生题目。
102. 二叉树的层序遍历
思路：
以前的笔记
代码：
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> list = new ArrayList<List<Integer>>();
if (root == null) {
return list;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while (!queue.isEmpty()) {
List<Integer> levelList = new LinkedList<>();
int len = queue.size();
// 遍历这一层
for (int i = 0; i < len; i++) {
TreeNode node = queue.poll();
levelList.add(node.val);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
list.add(levelList);
}
return list;
}
}226. 翻转二叉树
思路：
本题目可以使用递归，迭代和层序遍历来做。需要注意的是迭代的方法中，只能使用前序和后序遍历，因为使用中序遍历会对同一个子树翻转两次。
代码：
/**
* 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;
* }
* }
*/
// DFS递归方法
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return root;
}
swapChildren(root);
invertTree(root.left);
invertTree(root.right);
return root;
}
public void swapChildren(TreeNode node) {
TreeNode tem = node.left;
node.left = node.right;
node.right = tem;
}
}
// 迭代的前序遍历
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return root;
}
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
swapChildren(node);
if (node.right != null) {
stack.push(node.right);
}
if (node.left != null) {
stack.push(node.left);
}
}
return root;
}
public void swapChildren(TreeNode node) {
TreeNode tem = node.left;
node.left = node.right;
node.right = tem;
}
}
// 层序遍历
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return root;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
swapChildren(node);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
return root;
}
public void swapChildren(TreeNode node) {
TreeNode tem = node.left;
node.left = node.right;
node.right = tem;
}
}101. 对称二叉树
思路：
dfs：判断一个树是否对称，从根节点开始，看左子树和右子树是否对称。】
bfs：通过队列进行遍历比较，把左子树根节点和右子树根节点入队列，比较因素同dfs。需要注意的是，后续入队列时，左子树的左节点和右子树的右节点，左子树的右节点和右子树的左节点一起入队，两两入队列，两两出队列比较。
代码：
/**
* 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;
* }
* }
*/
// dfs
class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) {
return true;
}
return compare(root.left, root.right);
}
// 判断一个节点的左子树和右子树是否对称
public boolean compare(TreeNode left, TreeNode right) {
// 左子树和右子树都为空，对称
if (left == null && right == null) {
return true;
}
// 左右子树只有一个为空，不对称
if (left == null || right == null) {
return false;
}
// 左子树根节点值不等于右子树根节点值，不对称
if (left.val != right.val) {
return false;
}
// 只剩下left.val和right.val相等的情况
// 需要判断left.left和right.right, left.right和right.left是否对称
// boolean outside = compare(left.left, right.right);
// boolean inside = compare(left.right, right.left);
// return outside && inside;
// 简化
return compare(left.left, right.right) && compare(left.right, right.left);
}
}
// bfs
class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) {
return true;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root.right);
queue.offer(root.left);
while (!queue.isEmpty()) {
TreeNode right = queue.poll();
TreeNode left = queue.poll();
if (left == null && right == null) {
continue;
}
if (left == null || right == null) {
return false;
}
if (left.val != right.val) {
return false;
}
// left.val等于right.val
queue.offer(left.left);
queue.offer(right.right);
queue.offer(left.right);
queue.offer(right.left);
}
return true;
}
}
这两道题目基本和本题是一样的，只要稍加修改就可以AC。
- 100.相同的树(opens new window)
- 572.另一个树的子树
100. 相同的树
代码：
// dfs
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;
}
// p.val和q.val相等的情况
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}
// bfs
class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null) {
return true;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(p);
queue.offer(q);
while (!queue.isEmpty()) {
TreeNode left = queue.poll();
TreeNode right = queue.poll();
if (left == null && right == null) {
continue;
}
if (left == null || right == null) {
return false;
}
if (left.val != right.val) {
return false;
}
queue.offer(left.left);
queue.offer(right.left);
queue.offer(left.right);
queue.offer(right.right);
}
return true;
}
}
