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DeleteNodeInABST450.java

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/**

* 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:

*

* Search for a node to remove.

* 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

*

*/

/**

* Definition for a binary tree node.

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

* TreeNode(int x) { val = x; }

* }

*/

public class DeleteNodeInABST450 {

public TreeNode deleteNode(TreeNode root, int key) {

if (root == null) return null;

if (key < root.val) {

root.left = deleteNode(root.left, key);

return root;

}

if (key > root.val) {

root.right = deleteNode(root.right, key);

return root;

}

return delete(root);

}

private TreeNode delete(TreeNode node) {

if (node.left == null) return node.right;

if (node.right == null) return node.left;

return moveNext(node);

}

private TreeNode moveNext(TreeNode node) {

TreeNode min = node.right;

if (min.left == null) {

node.val = min.val;

node.right = min.right;

return node;

}

while (min.left != null && min.left.left != null) {

System.out.println(min.val);

min = min.left;

}

node.val = min.left.val;

min.left = delete(min.left);

return node;

}

public TreeNode deleteNode2(TreeNode root, int key) {

/* Base Case: If the tree is empty */

if (root == null) return root;

/* Otherwise, recur down the tree */

if (key < root.val) {

root.left = deleteNode(root.left, key);

} else if (key > root.val) {

root.right = deleteNode(root.right, key);

} else {

// node with only one child or no child

if (root.left == null) {

return root.right;

} else if (root.right == null) {

return root.left;

}

// node with two children: Get the inorder successor (smallest

// in the right subtree)

root.val = minValue(root.right);

// Delete the inorder successor

root.right = deleteNode(root.right, root.val);

}

return root;

}

private int minValue(TreeNode root) {

int minv = root.val;

while (root.left != null) {

minv = root.left.val;

root = root.left;

}

return minv;

}

}