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 v as 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.

4年之后，竟然做不出来...之前还不屑写笔记。其实还是能看出bfs。但如果insert到最后一行的时候，会加不上去因为叶子节点没有下一层。一看答案，原来是分两部分。先判断掉在根那一层加的。然后下面bfs，如果到了深度，我们就直接加上new node，不用入队，因为我们处理完这一层就ok了。记得先把level＋1，因为leve == d - 1好像最后一层不好处理。T:O(N), S:O(最宽一层宽度）

public TreeNode addOneRow(TreeNode root, int v, int d) {
    if (root == null || d < 1) {
        return null;
    }

    if (d == 1) {
        TreeNode newRoot = new TreeNode(v);
        newRoot.left = root;
        return newRoot;
    }

    Queue<TreeNode> queue = new LinkedList<>();
    queue.offer(root);
    int level = 1;
    while (!queue.isEmpty()) {
        int size = queue.size();
        // one level
        level++;
        for (int i = 0; i < size; i++) {
            TreeNode cur = queue.poll();

            if (d == level) {
                TreeNode newLeft = new TreeNode(v);    
                newLeft.left = cur.left;
                cur.left = newLeft;
                TreeNode newRight = new TreeNode(v);    
                newRight.right = cur.right;
                cur.right = newRight;
            } else {

                if (cur.left != null) {
                    queue.offer(cur.left);
                }

                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
        }
    }

    return root;
}