leetcode树专题894.897,919,951

满二叉树是一类二叉树,其中每个结点恰好有 0 或 2 个子结点。

返回包含 N 个结点的所有可能满二叉树的列表。 答案的每个元素都是一个可能树的根结点。

答案中每个树的每个结点都必须有 node.val=0

你可以按任何顺序返回树的最终列表。

示例:

输入:7
输出:[[0,0,0,null,null,0,0,null,null,0,0],[0,0,0,null,null,0,0,0,0],[0,0,0,0,0,0,0],[0,0,0,0,0,null,null,null,null,0,0],[0,0,0,0,0,null,null,0,0]]
解释:

提示:

  • 1 <= N <= 20

递归做的,

贴代码:

public List<TreeNode> allPossibleFBT(int N) {
        List<TreeNode>list=new ArrayList<TreeNode>();
         TreeNode node=new TreeNode(0);
        if(N==0)
            return null;
        if(N==1){
            list.add(node);
            return list;
        }

        else{
            List<TreeNode>list1=new ArrayList<TreeNode>();
            for(int h=1;h<N-1;h++){
             if(h%2==0)
                continue;
             List<TreeNode>left=allPossibleFBT(h);
             List<TreeNode>right=allPossibleFBT(N-h-1);
            if(h==3){
                    //System.out.println(right.size());
                    //System.out.println(right.get(0).left.val);
                     //System.out.println(right.get(0).right.right.val);
                    //System.out.println(right.get(1).left.left.val);
                     //System.out.println(right.get(1).right.val);
               }
             for(int j=0;j<left.size();j++){

                for(int i=0;i<right.size();i++){
                      TreeNode node1=new TreeNode(0);
                     node1.left=left.get(j);
                    node1.right=right.get(i);
                    list1.add(node1);
                }
            }
        }
            return list1;

    }

    }

艾玛这题有点坑,刚开始我把TreeNode初始化放在第一层循环里面,结果解答错误,放在第二层循环就对了,求好心人告知这是什么原因,像下面这样。我是怎么发现这个的呢。我是实在找不到bug了,让一个大佬看看,然后他觉得放在里面好看于是挪进去,于是就通过了,我去,惊了。

897递增顺序查找树

给定一个树,按中序遍历重新排列树,使树中最左边的结点现在是树的根,并且每个结点没有左子结点,只有一个右子结点。

示例 :

输入:[5,3,6,2,4,null,8,1,null,null,null,7,9]

       5
      /     3    6
   / \      2   4    8
 /        / \
1        7   9

输出:[1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

 1
     2
         3
             4
                 5
                     6
                         7
                             8
                                 9  

提示:

  1. 给定树中的结点数介于 1 和 100 之间。
  2. 每个结点都有一个从 0 到 1000 范围内的唯一整数值。

树的题目做多了,发现遍历很重要,前序、中序、后序和层次遍历都应该掌握

  public TreeNode increasingBST(TreeNode root) {
        Stack <TreeNode>stack=new Stack<TreeNode>();
        List<Integer>re=new ArrayList<Integer>();
        int tag=0;

        while(root!=null||!stack.isEmpty()){
            while(root!=null){
                stack.push(root);
                root=root.left;
            }
            if(!stack.isEmpty()){
                TreeNode node=stack.pop();
               re.add(node.val);
                root=node.right;
            }
        }
        TreeNode next=new TreeNode(re.get(re.size()-1));
        for(int i=re.size()-2;i>=0;i--){
           TreeNode node=new TreeNode(re.get(i));
            node.right=next;
            next=node;
        }
        return next;
    }

919. 完全二叉树插入器

这题还是层次遍历的变形,当然我的这种做法并不好,用了243Ms,感觉快要超时了

class CBTInserter {
   TreeNode root;
    public CBTInserter(TreeNode root) {
       this.root=root;
    }

    public int insert(int v) {
       Queue <TreeNode>queue=new LinkedList<TreeNode>();
        queue.offer(root);
        int re=0;
        int flag=0;
        while(!queue.isEmpty()){
            for(int i=0;i<queue.size();i++){
                TreeNode node=queue.poll();
                if(node.left==null&&node.right==null){
                    node.left=new TreeNode(v);
                    re=node.val;
                    flag=1;
                }
                else if(node.right==null){
                    node.right=new TreeNode(v);
                    re=node.val;
                    flag=1;
                }
                else{
                    queue.offer(node.left);
                    queue.offer(node.right);
                }
               if(flag==1)
                    break;

            }
             if(flag==1)
                    break;
        }
        return re;

    }

    public TreeNode get_root() {

     return root;
    }
}

951.

翻转等价二叉树

简单的递归,嘿嘿

 public boolean flipEquiv(TreeNode root1, TreeNode root2) {
        if(root1==null&&root2==null)
            return true;
        else if(root1==null||root2==null)
            return false;
        if(root1.val!=root2.val)
            return false;
        if(root1.left==null&&root1.right==null&&root2.left==null&&root2.right==null)
            return true;
        boolean judge1=flipEquiv(root1.left,root2.left)&&flipEquiv(root2.right,root1.right);
        boolean judge2=flipEquiv(root1.left,root2.right)&&flipEquiv(root1.right,root2.left);
        return judge1||judge2;

    }

原文地址:https://www.cnblogs.com/HannahLihui/p/10159347.html

时间: 2024-10-27 11:43:05

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