刷题进行时-递归-450. 删除二叉搜索树中的节点

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。


package 递归;

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode() {
    }

    TreeNode(int val) {
        this.val = val;
    }

    TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

class DeleteNodeinaBST450 {
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) {
            return null;
        }
        if (root.val > key) {
            root.left = deleteNode(root.left, key);
            return root;
        }
        if (root.val < key) {
            root.right = deleteNode(root.right, key);
            return root;
        }
        if (root.val == key) {
            if (root.left == null && root.right == null) {
                return null;
            }
            if (root.right == null) {
                return root.left;
            }
            if (root.left == null) {
                return root.right;
            }
            TreeNode successor = root.right;
            while (successor.left != null) {
                successor = successor.left;
            }
            root.right = deleteNode(root.right, successor.val);
            successor.right = root.right;
            successor.left = root.left;
            return successor;
        }
        return root;
    }

    public static void main(String[] args) {
        DeleteNodeinaBST450 so = new DeleteNodeinaBST450();
        TreeNode a5 = new TreeNode(5);
        TreeNode a2 = new TreeNode(2);
        TreeNode a3 = new TreeNode(3);
        TreeNode a4 = new TreeNode(4);
        TreeNode a6 = new TreeNode(6);
        TreeNode a7 = new TreeNode(7);
        a5.left = a3;
        a5.right = a6;
        a3.left = a2;
        a3.right = a4;
        a6.right = a7;
        TreeNode b = so.deleteNode(a5, 3);
        int c = 1;
    }
}