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'soriginal 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.
Constraints
The given d is in range [1, maximum depth of the given tree + 1].
The given binary tree has at least one tree node.
Approach
Links
GeeksforGeeks
ProgramCreek
YouTube
Examples
Input: [4, 2, 6, 3, 1, 5], v=1, d=2
Output: [4, 1, 1, 2, null, null, 6, 3, 1, 5]
Input: [4, 2, null, 3, 1], v=1, d=3
Output: [4, 2, null, 1, 1, 3, null, null, 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;
}
}
/**
* Time complexity : O(n). A total of n nodes of the given tree
* will be considered.
* Space complexity : O(n). The depth of the recursion tree can go upto n
* in the worst case(skewed tree).
*/
class Solution {
public TreeNode addOneRow(TreeNode root, int v, int d) {
if(d == 1) {
TreeNode node = new TreeNode(v);
node.left = root;
return node;
}
addOneRowHelper(root, v, d, 1);
return root;
}
private void addOneRowHelper(TreeNode root, int v, int depth, int currDepth) {
if(root == null) {
return;
}
if(currDepth < depth-1) {
addOneRowHelper(root.left, v, depth, currDepth+1);
addOneRowHelper(root.right, v, depth, currDepth+1);
}
if(currDepth == depth-1) {
TreeNode left = root.left;
TreeNode right = root.right;
root.left = new TreeNode(v);
root.right = new TreeNode(v);
root.left.left = left;
root.right.right = right;
}
}
}
/**
* Time complexity : O(n). A total of n nodes of the given tree
* will be considered.
* Space complexity : O(n). The depth of the recursion tree can go upto n
* in the worst case(skewed tree).
*/
public class Solution {
class Node{
Node(TreeNode n,int d){
node=n;
depth=d;
}
TreeNode node;
int depth;
}
public TreeNode addOneRow(TreeNode t, int v, int d) {
if (d == 1) {
TreeNode n = new TreeNode(v);
n.left = t;
return n;
}
Stack<Node> stack=new Stack<>();
stack.push(new Node(t,1));
while(!stack.isEmpty())
{
Node n=stack.pop();
if(n.node==null)
continue;
if (n.depth == d - 1 ) {
TreeNode temp = n.node.left;
n.node.left = new TreeNode(v);
n.node.left.left = temp;
temp = n.node.right;
n.node.right = new TreeNode(v);
n.node.right.right = temp;
} else{
stack.push(new Node(n.node.left, n.depth + 1));
stack.push(new Node(n.node.right, n.depth + 1));
}
}
return t;
}
}
/**
* Time complexity : O(n). A total of n nodes of the given tree will
* be considered in the worst case.
* Space complexity : O(x). The size of the queuequeue or temp queue can
* grow upto x only. Here, x refers to the number of maximum number
* of nodes at any level in the given tree.
*/
public class Solution {
public TreeNode addOneRow(TreeNode t, int v, int d) {
if (d == 1) {
TreeNode n = new TreeNode(v);
n.left = t;
return n;
}
Queue < TreeNode > queue = new LinkedList < > ();
queue.add(t);
int depth = 1;
while (depth < d - 1) {
Queue < TreeNode > temp = new LinkedList < > ();
while (!queue.isEmpty()) {
TreeNode node = queue.remove();
if (node.left != null) temp.add(node.left);
if (node.right != null) temp.add(node.right);
}
queue = temp;
depth++;
}
while (!queue.isEmpty()) {
TreeNode node = queue.remove();
TreeNode temp = node.left;
node.left = new TreeNode(v);
node.left.left = temp;
temp = node.right;
node.right = new TreeNode(v);
node.right.right = temp;
}
return t;
}
}
