[数据结构与算法]leetcode-删除二叉搜索树中的节点

一.题目描述

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。

https://leetcode-cn.com/problems/delete-node-in-a-bst/

二.代码

    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) {
            return root;
        }
        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;
            } else {
                TreeNode cur = root.right;
                while (cur.left != null) {
                    cur = cur.left;
                }
                cur.left = root.left;
                return root.right;
            }
        }
        return root;
    }

    public int successor(TreeNode root) {
        root = root.right;
        while (root.left != null) {
            root = root.left;
        }
        return root.val;
    }

    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;
        }
        if (key > root.val) {
            root.right = deleteNode(root.right, key);
        } else if (key < root.val) {
            root.left = deleteNode(root.left, key);
        } else {
            if (root.left == null && root.right == null) {
                root = null;
            } else if (root.right != null) {
                root.val = successor(root);
                root.right = deleteNode(root.right, root.val);
            } else {
                root.val = predecessor(root);
                root.left = deleteNode(root.left, root.val);
            }
        }
        return root;
    }