450 Delete Node in a BST

450. Delete Node in a BST

1. Question

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:

1. Search for a node to remove.

2. 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;
    }
}

3. Time & Space Complexity

Iteration:时间复杂度O(h), 空间复杂度O(1)

Recursion: 时间复杂度O(h), 空间复杂度O(h)