Leetcode 450. Delete Node in a BST 删除BST中的节点 解题报告

1 解题思想

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。

要保证整个性质，我们必须在删除的位置上，找一个合适的值来进行替换，使得BST上的每个节点都满足 当前节点的值大于左节点但是小于右节点

而替换策略就是：
1、当前删除位置，用左边子树的最大值的节点替换
2、或者是，用右边子树的最小值的节点替换

用上面的策略就可以保证删除后性质不变，并且调整开销也很少

2 原题

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:Search for a node to remove.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

3 AC解

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public int findReplacement(TreeNode parent,TreeNode node,boolean isLeft){        if(node.right == null){            if (isLeft)                parent.left = node.left;            else parent.right = node.left;            return node.val;        }        return findReplacement(node,node.right,false);    }    public TreeNode deleteNode(TreeNode root, int key) {        if(root==null)            return null;        if(root.val == key){            if(root.left == null)                return root.right;            if(root.right == null)                return root.left;            root.val = findReplacement(root,root.left,true); // 选择左边最大的，或者右边最小的        } else{            if(root.val > key)                root.left = deleteNode(root.left,key);            else                root.right = deleteNode(root.right,key);        }        return root;    }}