LeetCode 623. Add One Row to Tree

原题链接在这里：https://leetcode.com/problems/add-one-row-to-tree/

题目：

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.

Example 1:

Input: 
A binary tree as following:
       4
     /   \
    2     6
   / \   / 
  3   1 5   

v = 1

d = 2

Output: 
       4
      / \
     1   1
    /     \
   2       6
  / \     / 
 3   1   5   

 

Example 2:

Input: 
A binary tree as following:
      4
     /   
    2    
   / \   
  3   1    

v = 1

d = 3

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

Note:

1. The given d is in range [1, maximum depth of the given tree + 1].
2. The given binary tree has at least one tree node.

题解：

Could do BFS like level order traversal, get the queue of d-1 level nodes.

For each of them insert new node.

Corner case include when d == 1. Append current root to new root's left and return new root.

Time Complexity: O(n).

Space: O(n). Queue size.

AC Java:

/**
 * 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 {
    public TreeNode addOneRow(TreeNode root, int val, int depth) {
        if(root == null || depth < 1){
            return root;
        }

        if(depth == 1){
            TreeNode newRoot = new TreeNode(val);
            newRoot.left = root;
            return newRoot;
        }

        LinkedList<TreeNode> que = new LinkedList<>();
        que.add(root);
        while(!que.isEmpty() && depth > 2){
            int size = que.size();
            while(size-- > 0){
                TreeNode cur = que.poll();
                if(cur.left != null){
                    que.add(cur.left);
                }

                if(cur.right != null){
                    que.add(cur.right);
                }
            }

            depth--;
        }

        while(!que.isEmpty()){
            TreeNode cur = que.poll();
            TreeNode left = new TreeNode(val);
            left.left = cur.left;
            cur.left = left;

            TreeNode right = new TreeNode(val);
            right.right = cur.right;
            cur.right = right;
        }

        return root;
    }
}

Could also do DFS. DFS call need to know the current depth, when it is equal to d-1. Insert the new node.

Otherwise, keep doing DFS.

Time Complexity: O(n).

Space: (h).

AC Java:

/**
 * 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;
        }
        
        insert(root, v, d, 1);
        return root;
    }
    
    private void insert(TreeNode root, int v, int d, int curDepth){
        if(root == null){
            return;
        }
        
        if(curDepth == d-1){
            TreeNode left = new TreeNode(v);
            left.left = root.left;
            root.left = left;
            
            TreeNode right = new TreeNode(v);
            right.right = root.right;
            root.right = right;
        }else{
            insert(root.left, v, d, curDepth+1);
            insert(root.right, v, d, curDepth+1);
        }
    }
}