zoukankan      html  css  js  c++  java
  • LeetCode 623. Add One Row to Tree

    原题链接在这里:https://leetcode.com/problems/add-one-row-to-tree/

    题目:

    Given the root of a binary tree, then value v and depth d, you need to add a row of nodes with value v at the given depth d. The root node is at depth 1.

    The adding rule is: given a positive integer depth d, for each NOT null tree nodes N in depth d-1, create two tree nodes with value v as N's left subtree root and right subtree root. And N's original left subtree should be the left subtree of the new left subtree root, its original right subtree should be the right subtree of the new right subtree root. If depth d is 1 that means there is no depth d-1 at all, then create a tree node with value v as the new root of the whole original tree, and the original tree is the new root's left subtree.

    Example 1:

    Input: 
    A binary tree as following:
           4
         /   
        2     6
       /    / 
      3   1 5   
    
    v = 1
    
    d = 2
    
    Output: 
           4
          / 
         1   1
        /     
       2       6
      /      / 
     3   1   5   
    
    

    Example 2:

    Input: 
    A binary tree as following:
          4
         /   
        2    
       /    
      3   1    
    
    v = 1
    
    d = 3
    
    Output: 
          4
         /   
        2
       /     
      1   1
     /       
    3       1

    Note:

    1. The given d is in range [1, maximum depth of the given tree + 1].
    2. The given binary tree has at least one tree node.


    题解:

    Could do BFS like level order traversal, get the queue of d-1 level nodes.

    For each of them insert new node.

    Corner case include when d == 1. Append current root to new root's left and return new root.

    Time Complexity: O(n).

    Space: O(n). Queue size.

    AC Java:

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 class Solution {
    11     public TreeNode addOneRow(TreeNode root, int v, int d) {
    12         if(d == 1){
    13             TreeNode newRoot = new TreeNode(v);
    14             newRoot.left = root;
    15             return newRoot;
    16         }
    17         
    18         if(d < 1){
    19             return root;
    20         }
    21         
    22         if(root == null){
    23             return root;
    24         }
    25         
    26         LinkedList<TreeNode> que = new LinkedList<TreeNode>();
    27         que.add(root);
    28         int curCount = 1;
    29         int nextCount = 0;
    30         while(!que.isEmpty() && d>2){
    31             TreeNode cur = que.poll();
    32             curCount--;
    33             if(cur.left != null){
    34                 que.add(cur.left);
    35                 nextCount++;
    36             }
    37             
    38             if(cur.right != null){
    39                 que.add(cur.right);
    40                 nextCount++;
    41             }
    42             
    43             if(curCount == 0){
    44                 d--;
    45                 
    46                 curCount = nextCount;
    47                 nextCount = 0;
    48             }
    49         }
    50         
    51         while(!que.isEmpty()){
    52             TreeNode cur = que.poll();
    53             TreeNode left = new TreeNode(v);
    54             left.left = cur.left;
    55             cur.left = left;
    56             
    57             TreeNode right = new TreeNode(v);
    58             right.right = cur.right;
    59             cur.right = right;
    60         }
    61         
    62         return root;
    63     }
    64 }

    Could also do DFS. DFS call need to know the current depth, when it is equal to d-1. Insert the new node.

    Otherwise, keep doing DFS.

    Time Complexity: O(n).

    Space: (h).

    AC Java:

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 class Solution {
    11     public TreeNode addOneRow(TreeNode root, int v, int d) {
    12         if(d == 1){
    13             TreeNode newRoot = new TreeNode(v);
    14             newRoot.left = root;
    15             return newRoot;
    16         }
    17         
    18         insert(root, v, d, 1);
    19         return root;
    20     }
    21     
    22     private void insert(TreeNode root, int v, int d, int curDepth){
    23         if(root == null){
    24             return;
    25         }
    26         
    27         if(curDepth == d-1){
    28             TreeNode left = new TreeNode(v);
    29             left.left = root.left;
    30             root.left = left;
    31             
    32             TreeNode right = new TreeNode(v);
    33             right.right = root.right;
    34             root.right = right;
    35         }else{
    36             insert(root.left, v, d, curDepth+1);
    37             insert(root.right, v, d, curDepth+1);
    38         }
    39     }
    40 }
  • 相关阅读:
    php使用cookie来保存用户登录信息
    Linux下进程操作
    TortoiseSVN无法编辑日志信息的解决方法
    用DIV布局制作公告板
    HashMap的遍历
    实现文本滚动
    子DIV块中设置margintop时影响父DIV块位置的解决办法
    php使用session来保存用户登录信息
    javascript jquery ajax动态提交多个参数 api测试 拂晓风起
    javascript 处理返回json中的\u中文乱码问题(也不是乱码了,就是\u编码) 拂晓风起
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11100550.html
Copyright © 2011-2022 走看看