450. Delete Node in a BST

 

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.

Basically, the deletion can be divided into two stages:

  1. Search for a node to remove.
  2. If the node is found, delete the node.

 

Note: Time complexity should be O(height of tree).

Example:

root = [5,3,6,2,4,null,7]
key = 3

    5
   / \
  3   6
 / \   \
2   4   7

Given key to delete is 3. So we find the node with value 3 and delete it.

One valid answer is [5,4,6,2,null,null,7], shown in the following BST.

    5
   / \
  4   6
 /     \
2       7

Another valid answer is [5,2,6,null,4,null,7].

    5
   / \
  2   6
   \   \
    4   7

 

思路：

注意是BST 已经排好序了。

找到要删除的node;

node 不含左右节点  返回null;

node 只含有左子树，返回左子树

node只含有右子树，返回右子树

node 左右子树都有，找到右子树种最下的值，赋给node,递归地删掉 有字数中最小的元素

 

 1 class Solution {
 2     public TreeNode deleteNode(TreeNode root, int key) {
 3         if(root==null) return null ;
 4         if(root.val>key) root.left=deleteNode(root.left,key);
 5         else if(root.val<key) root.right = deleteNode(root.right,key);
 6         else {
 7             
 8             if(root.left==null) return root.right;
 9             if(root.right==null) return root.left;
10             
11             TreeNode rightmin = findmin(root.right);
12             root.val = rightmin.val;
13             root.right = deleteNode(root.right,root.val);
14         }
15         return root;
16     }
17     private TreeNode findmin(TreeNode root){
18         while(root.left!=null)
19             root=root.left;
20         return root;
21     }
22    
23     
24 }

 

 

 

 
posted @ 2017-11-17 10:51  乐乐章  阅读(358)  评论(0)  编辑  收藏  举报