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 }