Add One Row to Tree

Problem

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:

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

Solution

Recursive Solution: O(n) time, O(n) space

public class Solution {

    public TreeNode addOneRow(TreeNode root, int v, int d) {
        if (root == null || d < 1) return root; //check corner cases 
        if (d == 1) {
            TreeNode newRoot = new TreeNode(v);
            newRoot.left = root;
            return newRoot;
        }
        helper(root, v, d - 1);
        return root;
    }

    private void helper(TreeNode node, int v, int level) {
        if (level == 1) {
            TreeNode newLeft = new TreeNode(v), newRight = new TreeNode(v);
            newLeft.left = node.left;
            newRight.right = node.right;
            node.left = newLeft;
            node.right = newRight;
            return;
        }
        if (node.left != null) helper(node.left, v, level - 1);
        if (node.right != null) helper(node.right, v, level - 1);
    }
}

Iterative Solution: O(n) time, O(x) space, where x is the maximum nodes in any level

public class Solution {
    public TreeNode addOneRow(TreeNode root, int v, int d) {
        if (root == null || d < 1) return root;
        if (d == 1) {
            TreeNode newRoot = new TreeNode(v);
            newRoot.left = root;
            return newRoot;
        }
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int depth = 1;
        while (depth < d - 1) {
            Queue<TreeNode> temp = new LinkedList<>();
            while (!q.isEmpty()) {
                TreeNode cur = q.poll();
                if (cur.left != null) temp.offer(cur.left);
                if (cur.right != null) temp.offer(cur.right);
            }
            q = temp;
            depth++;
        }
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            TreeNode newLeft = new TreeNode(v), newRight = new TreeNode(v);
            newLeft.left = node.left;
            newRight.right = node.right;
            node.left = newLeft;
            node.right = newRight;
        }
        return root;
    } 
}

Analysis

The idea in both solutions is same
In the recursive solution, we update entire row by recursively calling the helper() method with node.left and node.right
In the iterative solution, we do that by using while loop while(!q.isEmpty())
To get every element in the same depth, we use temp queue and BFS approach
The inserting part is exactly same in both solutions
Where we first get current node and then construct newLeft and newRight with given v
Then we set their left and right with current node.left and node.right respectively
Then we set node.left and node.right with newLeft and newRight

Notice that the iterative solution use O(x) space because the size of q is decided based on temp
Where temp won't exceed the maximum number of nodes in any depth