LeetCode 450. Delete Node in a BST

原题链接在这里：https://leetcode.com/problems/delete-node-in-a-bst/?tab=Description

题目：

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

题解：

根据BST的特性找key的TreeNode node. 然后delete掉. 有几种情况:

找不到, return null.

node.left == null, return node.right.

node.right == null, return node.left.

左右都不为空，找node的successor, swap value, 然后delete掉successor.

Time Complexity: O(h). Space: O(h), stack sapce.

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if(root == null){
            return root;
        }
        if(key < root.val){
            root.left = deleteNode(root.left, key);
        }else if(key > root.val){
            root.right = deleteNode(root.right, key);
        }else{
            if(root.left == null){
                return root.right;
            }else if(root.right == null){
                return root.left;
            }else{
                TreeNode suc = findSuc(root.right);
                root.val = suc.val;
                root.right = deleteNode(root.right, suc.val);
            }
        }
        return root;
    }
    
    private TreeNode findSuc(TreeNode root){
        while(root.left != null){
            root = root.left;
        }
        return root;
    }
}

Iteration Method. 

在找key node时用pre来maintain 之前的位置, 若是找到的cur是pre的left, pre的left就更新为去掉cur后的子树. 反之亦然.

delete node时若是该node左右子树都不为null, 找successor然后swap. 返回successor.

Time Complexity: O(h). Space: O(1).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if(root == null){
            return root;
        }
        TreeNode cur = root;
        TreeNode pre = null;
        while(cur!=null && cur.val!=key){
            pre = cur;
            if(key < cur.val){
                cur = cur.left;
            }else if(key > cur.val){
                cur = cur.right;
            }
        }
        if(pre == null){
            return delete(cur);
        }
        if(pre.left == cur){
            pre.left = delete(cur);
        }else{
            pre.right = delete(cur);
        }
        return root;
    }
    
    private TreeNode delete(TreeNode root){
        if(root == null){
            return null;
        }else if(root.left == null){
            return root.right;
        }else if(root.right == null){
            return root.left;
        }
        TreeNode suc = root.right;
        TreeNode pre = null;
        while(suc.left != null){
            pre = suc;
            suc = suc.left;
        }
        suc.left = root.left;
        if(root.right != suc){
            pre.left = suc.right;
            suc.right = root.right;
        }
        return suc;
    }
}

类似Split BST.