Add One Row to Tree

Published: Oct 5, 2022

Medium Breadth-First Search Depth-First Search Binary Tree

Problem 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.

The adding rule is:

  • 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.

Constraints:

  • The number of nodes in the tree is in the range [1, 10**4].
  • The depth of the tree is in the range [1, 10**4].
  • -100 <= Node.val <= 100
  • -10**5 <= val <= 10**5
  • 1 <= depth <= the depth of tree + 1

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

Examples

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]
Explanation:
Input:
      4
    /   \
  2      6
 / \    /
3   1   5

Output:
       4
     /   \
    1     1
   /       \
  2        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]
Explanation:
Input:
     4
    /
  2
 / \
3   1

Output:
      4
     /
    2
   /  \
  1    1
 /      \
3        1

How to Solve

Both of depth-first search with/without stack and the breadth-first search work. Once the level to add new nodes is found, just create two new nodes and set those in the tree.

We should be careful when depth is 1. It needs a special treatment. The solution taken here is the breadth-first search approach. The node and depth pair is saved in a queue. When the current depth - 1 is the given depth, create two new nodes and add those to the current parent. The new nodes’ children are current parent’s children.

Solution

/**
 * 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:
    TreeNode* addOneRow(TreeNode* root, int val, int depth) {
        if (depth == 1) {
            return new TreeNode(val, root, nullptr);
        }
        queue<pair<TreeNode*, int>> q;
        q.push({root, 1});
        while (!q.empty()) {
            auto [cur, d] = q.front();
            q.pop();
            if (d == depth - 1) {
                TreeNode* left = cur->left;
                TreeNode* right = cur->right;
                cur->left = new TreeNode(val, left, nullptr);
                cur->right = new TreeNode(val, nullptr, right);
            } else {
                if (cur->left) {
                    q.push({cur->left, d + 1});
                }
                if (cur->right) {
                    q.push({cur->right, d + 1});
                }
            }
        }
        return root;
    }
};


# 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:
    def addOneRow(self, root: Optional[TreeNode], val: int, depth: int) -> Optional[TreeNode]:
        if depth == 1:
            return TreeNode(val=val, left=root)
        q = [(root, 1)]
        while q:
            cur, d = q.pop(0)
            if d == depth - 1:
                left, right = cur.left, cur.right
                cur.left = TreeNode(val, left, None)
                cur.right = TreeNode(val, None, right)
            else:
                if cur.left:
                    q.append((cur.left, d + 1))
                if cur.right:
                    q.append((cur.right, d + 1))
        return root

Complexities

  • Time: O(n)
  • Space: O(k) – k: number of the nodes in the same level