Add One Row to Tree

https://leetcode.com/problems/add-one-row-to-tree/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.

Thoughts

看到在某一层加结点，自然想到bfs到那层。

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
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> queue = new LinkedList<>();
        queue.add(root);
        int depth = 0;
        while (!queue.isEmpty()) {
            depth++;
            int length = queue.size();
            if (depth == d - 1) {
                for (int i = 0; i < length; i++) {
                    TreeNode node = queue.remove();
                    TreeNode newLeft = new TreeNode(v);
                    TreeNode newRight = new TreeNode(v);
                    newLeft.left = node.left;
                    newRight.right = node.right;
                    node.left = newLeft;
                    node.right = newRight;
                }
                return root;
            }
            for (int i = 0; i < length; i++) {
                TreeNode node = queue.remove();
                if (node.left != null) queue.add(node.left);
                if (node.right != null) queue.add(node.right);
            }            
        }

        return root;
    }
}

Analysis

完成时间1刻钟。

Errors:

1. depth == d

2. 没考虑d == 1

时间复杂度为O(n).