LintCode-Remove node in Binary Search Tree

Given a root of Binary Search Tree with unique value for each node.  Remove the node with given value. If there is no such a node with given value in the binary search tree, do nothing. You should keep the tree still a binary search tree after removal.

Example

Given binary search tree:

          5

       /    \

    3          6

 /    \

2       4

Remove 3, you can either return:

          5

       /    \

    2          6

      \

         4

or :

          5

       /    \

    4          6

 /   

2

Analysis:

We should consider every possible situation carefully.

Solution:

 

 1 /**
 2  * Definition of TreeNode:
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left, right;
 6  *     public TreeNode(int val) {
 7  *         this.val = val;
 8  *         this.left = this.right = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     /**
14      * @param root: The root of the binary search tree.
15      * @param value: Remove the node with given value.
16      * @return: The root of the binary search tree after removal.
17      */
18     public TreeNode removeNode(TreeNode root, int value) {
19         if (root==null) return null;
20 
21         TreeNode preRoot = new TreeNode(0);
22         preRoot.left = root;
23         searchNodeRecur(preRoot,root,value);
24         return preRoot.left;
25     }
26 
27     public void searchNodeRecur(TreeNode pre, TreeNode cur, int val){
28         //if there is no such node, then do nothing.
29         if (cur==null) return;
30 
31         //find the node.
32         if (cur.val == val){
33             removeNode(pre,cur);
34             return;
35         } else if (cur.val > val){
36             searchNodeRecur(cur,cur.left,val);
37         } else 
38             searchNodeRecur(cur,cur.right,val);
39 
40     }
41 
42     public void removeNode(TreeNode pre, TreeNode cur){
43         //a leaf node.
44         if (cur.left==null && cur.right==null)
45             if (pre.left==cur) pre.left = null;
46             else pre.right = null;
47         else if (cur.left==null || cur.right==null){ //cur only has one sub tree.
48             TreeNode child = (cur.left==null) ? cur.right : cur.left;
49             if (pre.left==cur) pre.left = child;
50             else pre.right=child;
51         } else {    //has two sub trees, select the largest node in the left subtree
52             TreeNode pre2 = cur;
53             TreeNode cur2 = cur.left;
54             while (cur2.right!=null){
55                 pre2 = cur2;
56                 cur2 = cur2.right;
57             }
58             //a leaf node
59             if (cur2.left==null){
60                 if (pre2.left==cur2) pre2.left = null;
61                 else pre2.right=null;
62                 cur2.left = cur.left;
63                 cur2.right = cur.right;
64                 if (pre.left==cur) pre.left=cur2;
65                 else pre.right=cur2;
66             } else {
67                 if (pre2.left==cur2) pre2.left = cur2.left;
68                 else pre2.right = cur2.left;
69                 cur2.left = cur.left;
70                 cur2.right = cur.right;
71                 if (pre.left==cur) pre.left=cur2;
72                 else pre.right=cur2;
73             }
74         }
75     }
76 
77                 
78         
79 }

 

 

 

posted @ 2014-12-30 07:19  LiBlog  阅读(516)  评论(0)  编辑  收藏  举报