[Lintcode] Remove Node in Binary Search Tree

Remove Node in Binary Search Tree

Given a root of Binary Search Tree with the 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, does 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

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Lintcode Copyright Binary Search Tree Ah, first to a lazy writing it.  There are too many categories of discussions. T. T

1 /**2 * Definition of TreeNode:3 * Class TreeNode {4 * Public:5 * int val;6 * TreeNode *left, *right;7 * TreeNode (int val) {8 * This->val = val;9 * This->left = This->right = NULL;Ten  *     } One  * } A  */ - classSolution { -  Public: the     /** - * @param root:the root of the binary search tree. - * @param value:remove the node with given value. - * @return: The root of the binary search tree after removal. +      */ -treenode* Buildtree (vector<treenode*> &v,intLeftintRight ) { +         if(Left > right)returnNULL; A         intMid = left + ((right-left) >>1); atV[mid]->left = Buildtree (V, left, mid-1); -V[mid]->right = Buildtree (V, mid +1, right); -         returnV[mid]; -     } -Treenode* RemoveNode (treenode* root,intvalue) { -         //Write your code here inVector<treenode*>v; -TreeNode *cur = root, *tmp; toStack<treenode*>Stk; +          while(cur! = NULL | |!Stk.empty ()) { -             if(cur! =NULL) { the Stk.push (cur); *Cur = cur->Left ; $}Else {Panax NotoginsengCur =stk.top (); - Stk.pop (); theTMP =cur; +Cur = cur->Right ; A                 if(Tmp->val! =value) { the V.push_back (TMP); +}Else { -                     Deletetmp; $TMP =NULL; $                 } -             } -         } the         returnBuildtree (V,0, (int) V.size ()-1); -     }Wuyi};

[Lintcode] Remove Node in Binary Search Tree