Java集合与数据结构——二叉树03

RAIN7

发布于 2022-03-31 13:51:22

1010

发布于 2022-03-31 13:51:22

文章被收录于专栏：RAIN7 de 编程之路

文章目录

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 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

/**
 * 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

    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。

 /**
 * 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.

    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;
    }