450. Delete Node in a BST

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

 含义：现在有一个二叉搜索树，现在要让你删除一个节点，并且保证整个BST的性质不变

 

/**
 * 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(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;
        }
        //右子树中最小值挪到当前root上，才能保持二叉树性质不变
        TreeNode minNode = findMin(root.right);
        root.val = minNode.val;
        root.right = deleteNode(root.right, root.val);
    }
    return root;
}

private TreeNode findMin(TreeNode node){
    while(node.left != null){
        node = node.left;
    }
    return node;
}
}