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
2. Implementation
(1) Iteration
class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
TreeNode preNode = null, curNode = root;
while (curNode != null && curNode.val != key) {
preNode = curNode;
if (curNode.val < key) {
curNode = curNode.right;
}
else{
curNode = curNode.left;
}
}
if (preNode == null) {
return getSuccessor(curNode);
}
else if (preNode.left == curNode) {
preNode.left = getSuccessor(curNode);
}
else if (preNode.right == curNode) {
preNode.right = getSuccessor(curNode);
}
return root;
}
public TreeNode getSuccessor(TreeNode node) {
if (node == null) {
return null;
}
if (node.left == null) {
return node.right;
}
if (node.right == null) {
return node.left;
}
TreeNode preNode = null, successor = node.right;
while (successor.left != null) {
preNode = successor;
successor = successor.left;
}
successor.left = node.left;
if (node.right != successor) {
preNode.left = successor.right;
successor.right = node.right;
}
return successor;
}
}
（2) Recursion
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
if (root == null) {
return null;
}
if (root.val < key) {
root.right = deleteNode(root.right, key);
}
else if (root.val > key) {
root.left = deleteNode(root.left, key);
}
else {
if (root.left == null) {
return root.right;
}
if (root.right == null) {
return root.left;
}
TreeNode successor = getInorderSuccessor(root.right);
// 注意这一步是改变了原来的数据，面试时不一定被允许
root.val = successor.val;
root.right = deleteNode(root.right, root.val);
}
return root;
}
public TreeNode getInorderSuccessor(TreeNode node) {
while (node.left != null) {
node = node.left;
}
return node;
}
}