【树】450. 删除二叉搜索树中的节点 (BST)应用BST的三个性质(复习)
题目:
给定一个二叉搜索树的根节点 root 和一个值 key,删除二叉搜索树中的 key 对应的节点,并保证二叉搜索树的性质不变。返回二叉搜索树(有可能被更新)的根节点的引用。
一般来说,删除节点可分为两个步骤:
首先找到需要删除的节点;
如果找到了,删除它。

解答:
class Solution { /* One step right and then always left */ public int successor(TreeNode root) { root = root.right; while (root.left != null) root = root.left; return root.val; } /* One step left and then always right */ public int predecessor(TreeNode root) { root = root.left; while (root.right != null) root = root.right; return root.val; } public TreeNode deleteNode(TreeNode root, int key) { if (root == null) return null; // delete from the right subtree if (key > root.val) root.right = deleteNode(root.right, key); // delete from the left subtree else if (key < root.val) root.left = deleteNode(root.left, key); // delete the current node else { // the node is a leaf if (root.left == null && root.right == null) root = null; // the node is not a leaf and has a right child else if (root.right != null) { root.val = successor(root); root.right = deleteNode(root.right, root.val); } // the node is not a leaf, has no right child, and has a left child else { root.val = predecessor(root); root.left = deleteNode(root.left, root.val); } } return root; } }
https://leetcode-cn.com/problems/delete-node-in-a-bst/solution/shan-chu-er-cha-sou-suo-shu-zhong-de-jie-dian-by-l/

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