文章目录
- 对称二叉树
- 二叉树的镜像
- 二叉树的层序遍历
- 01
- 02
- 判断一个二叉树是否为完全二叉树
- 二叉树的拓展问题
对称二叉树
代码语言:javascript复制/**
* 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 isSymmetricChild(TreeNode leftTree,TreeNode rightTree){
if(leftTree == null && rightTree != null){
return false;
}
if(leftTree != null && rightTree == null){
return false;
}
if(leftTree == null && rightTree == null){
return true;
}
if(leftTree.val != rightTree.val){
return false;
}
return isSymmetricChild(leftTree.left,rightTree.right)
&& isSymmetricChild(leftTree.right,rightTree.left);
}
public boolean isSymmetric(TreeNode root) {
if(root == null){
return true;
}
return isSymmetricChild(root.left,root.right);
}
}
二叉树的镜像
代码语言:javascript复制/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode mirrorTree(TreeNode root) {
if(root == null){
return root;
}
if(root.left == null && root.right == null){
return root;
}
TreeNode tmp = root.left;
root.left = root.right;
root.right = tmp;
if(root.left != null){
mirrorTree(root.left);
}
if(root.right != null){
mirrorTree(root.right);
}
return root;
}
}
二叉树的层序遍历
01
代码语言:javascript复制 void levelOrderTravelsal(TreeNode root){
if(root == null){
return ;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()){
TreeNode top = queue.poll();
System.out.print(top.val " ");
if(top.left != null){
queue.offer(top.left);
}
if(top.right != null){
queue.offer(top.right);
}
}
System.out.println();
}
}
实现效果
02
我们在这里看到的层序遍历还和之前我们做的不一样,我们来看结果,发现二叉树的每一层分别进行分行输出。
代码语言:javascript复制 /**
* 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 链表的作用只是为了满足结果的打印要求
List<List<Integer>> ret = new ArrayList<>();
if(root == null){
return ret ;
}
// 队列的作用是用来实现层序遍历的
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()){
int size = queue.size();
List<Integer> list = new ArrayList<>();
while(size !=0){
TreeNode top = queue.poll();
lists.add(top.val);
if(top.left != null){
queue.offer(top.left);
}
if(top.right != null){
queue.offer(top.right);
}
size--;
}
ret.add(list);
}
return ret;
}
}
判断一个二叉树是否为完全二叉树
利用之前的层序遍历,一层一层的遍历,当 top 为null 时,遍历停止,之后如果之后的队列中元素全部都是 null 的话 则返回true,这棵树是完全二叉树,如果这是队列中有一个元素不是 null,则返回false,这棵树不是完全二叉树.
代码语言:javascript复制 public boolean isCompleteTree(TreeNode root) {
if(root == null){
return true;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()){
TreeNode top = queue.poll();
if(top != null){
queue.offer(top.left);
queue.offer(top.right);
}else{
break;
}
}
while(!queue.isEmpty()){
TreeNode cur = queue.peek();
if(cur == null){
queue.poll();
}else{
return false;
}
}
return true;
}