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

    xiaoxiao2021-03-26  11

    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 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

    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; } }
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