LeetCode-in-Java

623. Add One Row to Tree

Medium

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.

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

Solution

import com_github_leetcode.TreeNode;

/*
 * 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;
 *     }
 * }
 */
public class Solution {
    public TreeNode addOneRow(TreeNode root, int val, int depth) {
        if (depth == 1) {
            TreeNode newRoot = new TreeNode(val);
            newRoot.left = root;
            return newRoot;
        }
        dfs(root, depth - 2, val);
        return root;
    }

    private void dfs(TreeNode node, int depth, int val) {
        if (depth == 0) {
            TreeNode left = new TreeNode(val);
            TreeNode right = new TreeNode(val);
            left.left = node.left;
            right.right = node.right;
            node.left = left;
            node.right = right;
        } else {
            if (node.left != null) {
                dfs(node.left, depth - 1, val);
            }
            if (node.right != null) {
                dfs(node.right, depth - 1, val);
            }
        }
    }
}

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