LeetCode[450]Delete Node in a BST(Java)

Description:

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   7Given 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       7Another valid answer is [5,2,6,null,4,null,7].    5   / \  2   6   \   \    4   7

Solution:

Recursive递归

首先查找到val=key节点的所在位置，在查找的过程中，利用递归重建BST。

查找到之后，需要将当前节点重新赋值，然后重新构建右子树，方式是将右子树的最左叶结点的值付给当前节点，然后删除该最左叶结点。

/** * 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.left = deleteNode(root.left, key);        }else if(root.val < key){            root.right = deleteNode(root.right, key);        }else{            if(root.left == null){                return root.right;            }else if(root.right == null){                return root.left;            }                        TreeNode temp = search(root.right);            root.val = temp.val;            root.right = deleteNode(root.right, temp.val);        }        return root;    }    public TreeNode search(TreeNode node){        while(node.left != null){            node = node.left;        }        return node;    }}