leetcode 每日一题 450. 删除二叉搜索树中的节点

leetcode 每日一题 450. 删除二叉搜索树中的节点

class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if(root == null){
            return null;
        }
        if(key == root.val){
            if(root.left != null){
                TreeNode rootLeftLastRight = right(root.left);
                rootLeftLastRight.right = root.right;
                return root.left;
            }else if(root.right != null){
                return root.right;
            }else {
                return null;
            }
        }
        f(root,key,null,null);
        return root;
    }

    private void f(TreeNode root, int key,TreeNode parent,Boolean b) {
        if (root == null){
            return;
        }
        //如果是这个要删除的节点
        if(root.val == key){
            //判断是父节点的左子节点还是右子节点
            if(b){
                //右边
                if(root.right != null){
                    //循环找到最后一个节点,然后把左的节点挂过去
                    TreeNode rootRightLastLeft = left(root.right);
                    rootRightLastLeft.left = root.left;
                    parent.right = root.right;
                }else if(root.left != null){
                    parent.right = root.left;
                }else {
                    parent.right = null;
                }
            }else{
                //左边
                if(root.left != null){
                    //循环找到最后一个节点,然后把右边的节点挂过去
                    TreeNode rootLeftLastRight = right(root.left);
                    rootLeftLastRight.right = root.right;
                    parent.left = root.left;
                }else if(root.right != null){
                    parent.left = root.right;
                }else {
                    parent.left = null;
                }
            }
        }
        f(root.left,key,root,false);
        f(root.right,key,root,true);
    }

    private TreeNode left(TreeNode root) {
        if(root.left == null){
            return root;
        }
        return left(root.left);
    }

    private TreeNode right(TreeNode root) {
        if(root.right == null){
            return root;
        }
        return right(root.right);
    }
}

改了几次，原因是左右节点的逻辑差不多，所以写的时候复制过去的，然后把right改成left，有部分代码没修改，导致一直测试出错