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.

class Solution {
    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.offer(root);
        int start = 1;
        while(!q.isEmpty()) {
            start++;
            if(start == d) {
                int size = q.size();
              for(int i = 0; i < size; i++) {
                  TreeNode cur = q.poll();
                  TreeNode tmp = cur.left;
                  cur.left = new TreeNode(v);
                  cur.left.left = tmp;
                  tmp = cur.right;
                  cur.right = new TreeNode(v);
                  cur.right.right = tmp;
              }
                break;
            }
            
            else {
                int size = q.size();
               for(int i = 0; i < size; i++) {
                    TreeNode cur = q.poll();
                    if(cur.left != null) q.offer(cur.left);
                    if(cur.right != null) q.offer(cur.right); 
                }
            }
        }
        return root;
    }
}

用bfs做，注意start起始要++，而且q.size()一直在变，所以不能用在for循环的终止条件里，先int size = q.size()比较保险。

class Solution {
    public TreeNode addOneRow(TreeNode root, int v, int d) {
        if(d == 1) {
            TreeNode cur = new TreeNode(v);
            cur.left = root;
            return cur;
        }
        Queue<TreeNode> q = new LinkedList();
        q.offer(root);
        int dep = 1;
        while(!q.isEmpty()) {
            dep += 1;
            if(dep == d) {
                int size = q.size();
                for(int i = 0; i < size; i++) {
                    TreeNode cur = q.poll();
                    TreeNode tmpl = new TreeNode(v);
                    tmpl.left = cur.left;
                    cur.left = tmpl;
                    TreeNode tmpr = new TreeNode(v);
                    tmpr.right = cur.right;
                    cur.right = tmpr;
                }
                break;
            }
            else {
                int size = q.size();
                for(int i = 0; i < size; i++) {
                    TreeNode cur = q.poll();
                    if(cur.left != null) q.offer(cur.left);
                    if(cur.right != null) q.offer(cur.right);
                }
            }
        }
        return root;
    }
}

用bfs是因为queue里存的是上一层，所以可以对当前一层进行修改