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: 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
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

思路

首先我们需要搜索这棵树，找到待删除节点的父亲节点。利用二叉搜索树的条件，可以对树进行二分搜索，搜索的复杂度就是树高了。

我们需要对节点进行删除操作，判断条件实际上就是三种情况：待删除的节点没有左右子树，有左/右子树，有左右子树。特别注意的是，有左右子树的情况，需要找到比待删除节点刚好大一点的数，也就是待删除节点右节点的最左端的节点。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
/**
 * 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 searchNode(TreeNode node, int key){
        if(node.left == null && node.right == null) return null;
        else{
            int leftval = -1;
            int rightval = -1;
            if(node.left != null) leftval = node.left.val;
            if(node.right != null) rightval = node.right.val;
            if(key == leftval || key == rightval) return node;
            else{
                if(key < node.val) return searchNode(node.left, key);
                else return searchNode(node.right, key);
            }
        }
    }
    public TreeNode deleteNode(TreeNode root, int key) {
        if(root == null) return root;
        TreeNode node = root;
        TreeNode parent = null;
        Boolean left = true;
        if(root.val != key){
            parent = searchNode(root, key);
            if(parent == null) return root;
            if(parent.left != null) if(parent.left.val == key) node = parent.left;
            if(parent.right != null) {
                if(parent.right.val == key){ 
                    node = parent.right;
                    left = false;
                }
            }
        }
        if(node.left == null && node.right == null){
            if(parent == null)  return null;
            if(left) parent.left = null;
            else parent.right = null;
        }
        else if(node.right != null && node.left == null){
            if(parent == null)  return node.right;
            else{
                if(left) parent.left = node.right;
                else parent.right = node.right;
            }
        }
        else if(node.right == null && node.left != null){
            if(parent == null) return node.left;
            else{
                if(left) parent.left = node.left;
                else parent.right = node.left;
            }
        }
        else{
            if(node.right != null){
                TreeNode leftmost = node.right;
                TreeNode leftmost_parent = node;
                while(leftmost.left != null){
                    leftmost_parent = leftmost;
                    leftmost = leftmost.left;
                }
                node.val = leftmost.val;
                if(leftmost_parent == node) node.right = node.right.right;
                else leftmost_parent.left = leftmost.right;
            }
            else{
                node.right = null;
            }
        }
        return root;
    }
}