623. Add One Row to Tree

Given the root of a binary tree, then value v and depth d, you need to add a row of nodes with value v at the given depth d. The root node is at depth 1.

The adding rule is: given a positive integer depth d, for each NOT null tree nodes N in depth d-1, create two tree nodes with value vas N's left subtree root and right subtree root. And N's original left subtree should be the left subtree of the new left subtree root, its original right subtree should be the right subtree of the new right subtree root. If depth d is 1 that means there is no depth d-1 at all, then create a tree node with value v as the new root of the whole original tree, and the original tree is the new root's left subtree.

Example 1:

Input: 
A binary tree as following:
       4
     /   
    2     6
   /    / 
  3   1 5   

v = 1

d = 2

Output: 
       4
      / 
     1   1
    /     
   2       6
  /      / 
 3   1   5   

Example 2:

Input: 
A binary tree as following:
      4
     /   
    2    
   /    
  3   1    

v = 1

d = 3

Output: 
      4
     /   
    2
   /     
  1   1
 /       
3       1

Note:

1. The given d is in range [1, maximum depth of the given tree + 1].
2. The given binary tree has at least one tree node.

题目含义：给二叉树增加一行，给了我们需要增加的值，还有需要增加的位置深度，题目中给的例子也比较能清晰的说明问题。但是漏了一种情况，那就是当d=1时，这该怎么加？这时候就需要替换根结点了

    public TreeNode addOneRow(TreeNode root, int v, int d) {
        if (d == 1) {
            TreeNode newroot = new TreeNode(v);
            newroot.left = root;
            return newroot;
        }

        Queue<TreeNode> q = new LinkedList<>();
        q.add(root);
        
        //找到要插入节点层的上一层
        for (int i = 0; i < d - 2; i++) {
            int size = q.size();
            for (int j = 0; j < size; j++) {
                TreeNode t = q.poll();
                if (t.left != null) q.add(t.left);
                if (t.right != null) q.add(t.right);
            }
        }
        while (!q.isEmpty()) {
            TreeNode t = q.poll();
            
            TreeNode tmp = t.left; //在左子树中插入行节点
            t.left = new TreeNode(v);
            t.left.left = tmp;

            tmp = t.right;//在右子树中插入行节点
            t.right = new TreeNode(v);
            t.right.right = tmp;
        }
        return root;        
    }