# 623. Add One Row to Tree

## Description

Given the root of a binary tree and two integers val and depth, add a row of nodes with value val at the given depth depth.

Note that the root node is at depth 1.

• Given the integer depth, for each not null tree node cur at the depth depth - 1, create two tree nodes with value val as cur's left subtree root and right subtree root.
• cur's original left subtree should be the left subtree of the new left subtree root.
• cur's original right subtree should be the right subtree of the new right subtree root.
• If depth == 1 that means there is no depth depth - 1 at all, then create a tree node with value val as the new root of the whole original tree, and the original tree is the new root's left subtree.

Example 1:

Input: root = [4,2,6,3,1,5], val = 1, depth = 2
Output: [4,1,1,2,null,null,6,3,1,5]


Example 2:

Input: root = [4,2,null,3,1], val = 1, depth = 3
Output: [4,2,null,1,1,3,null,null,1]


Constraints:

• The number of nodes in the tree is in the range [1, 104].
• The depth of the tree is in the range [1, 104].
• -100 <= Node.val <= 100
• -105 <= val <= 105
• 1 <= depth <= the depth of tree + 1

## Solutions

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private int val;
private int depth;

public TreeNode addOneRow(TreeNode root, int val, int depth) {
if (depth == 1) {
return new TreeNode(val, root, null);
}
this.val = val;
this.depth = depth;
dfs(root, 1);
return root;
}

private void dfs(TreeNode root, int d) {
if (root == null) {
return;
}
if (d == depth - 1) {
TreeNode l = new TreeNode(val, root.left, null);
TreeNode r = new TreeNode(val, null, root.right);
root.left = l;
root.right = r;
return;
}
dfs(root.left, d + 1);
dfs(root.right, d + 1);
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int val;
int depth;

TreeNode* addOneRow(TreeNode* root, int val, int depth) {
if (depth == 1) return new TreeNode(val, root, nullptr);
this->val = val;
this->depth = depth;
dfs(root, 1);
return root;
}

void dfs(TreeNode* root, int d) {
if (!root) return;
if (d == depth - 1) {
auto l = new TreeNode(val, root->left, nullptr);
auto r = new TreeNode(val, nullptr, root->right);
root->left = l;
root->right = r;
return;
}
dfs(root->left, d + 1);
dfs(root->right, d + 1);
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
self, root: Optional[TreeNode], val: int, depth: int
) -> Optional[TreeNode]:
def dfs(root, d):
if root is None:
return
if d == depth - 1:
root.left = TreeNode(val, root.left, None)
root.right = TreeNode(val, None, root.right)
return
dfs(root.left, d + 1)
dfs(root.right, d + 1)

if depth == 1:
return TreeNode(val, root)
dfs(root, 1)
return root


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func addOneRow(root *TreeNode, val int, depth int) *TreeNode {
if depth == 1 {
return &TreeNode{Val: val, Left: root}
}
var dfs func(root *TreeNode, d int)
dfs = func(root *TreeNode, d int) {
if root == nil {
return
}
if d == depth-1 {
l, r := &TreeNode{Val: val, Left: root.Left}, &TreeNode{Val: val, Right: root.Right}
root.Left, root.Right = l, r
return
}
dfs(root.Left, d+1)
dfs(root.Right, d+1)
}
dfs(root, 1)
return root
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function addOneRow(root: TreeNode | null, val: number, depth: number): TreeNode | null {
if (depth === 1) {
return new TreeNode(val, root);
}

const queue = [root];
for (let i = 1; i < depth - 1; i++) {
const n = queue.length;
for (let j = 0; j < n; j++) {
const { left, right } = queue.shift();
left && queue.push(left);
right && queue.push(right);
}
}
for (const node of queue) {
node.left = new TreeNode(val, node.left);
node.right = new TreeNode(val, null, node.right);
}
return root;
}