Description
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.

Post-order DFS in Java

Post-order traversing the tree, when it reaches the d-1 level of the tree, adding new nodes according to v, and then adding the original left/right nodes to the newly created nodes as children.
Recurssively doing this is trivial. Try iterative solution if you can.


/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**
    * @timeComplexity: O(n)
    * @runtime:        11ms
    * @algorithm:      Post-order DFS
    */
    public TreeNode addOneRow(TreeNode root, int v, int d) {
        if (d == 1) { // edge case
            TreeNode newRoot = new TreeNode(v);
            newRoot.left = root;
            return newRoot;
        }
        
        return dfs(root, 1, v, d);
        
    }
    
    private TreeNode dfs(TreeNode root, int level, int v, int d) {
        if (root == null)   return null;
        TreeNode leftNode = dfs(root.left, level+1, v, d);
        TreeNode rightNode = dfs(root.right, level+1, v, d);
        
        if (level == d-1) { // inserting a new level
            TreeNode newLeft = new TreeNode(v);
            TreeNode newRight = new TreeNode(v);
            newLeft.left = leftNode;
            newRight.right = rightNode;
            root.left = newLeft;
            root.right = newRight;
            return root;
        }
            
        return root;
    }
    
    
}

imgRuntime: 13ms