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LeetCode Weekly Contest 623.Add One Row to Tree

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Question

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: 

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18
       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: 

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      4
     /   
    2    
   / \   
  3   1    
v = 1
d = 3
Output:
      4
     /   
    2
   / \    
  1   1
 /     \  
3       1

Explanation

用BFS层级遍历，找到上一层以后，指定一下左右即可。还是要熟练掌握BFS。

Code

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public class Solution {
    public TreeNode addOneRow(TreeNode root, int v, int d) {
        if(d == 1) {
            TreeNode new_root = new TreeNode(v);
            new_root.left = root;
            return new_root;
        }
        // 存之前的每一层，BFS
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int level = 1;
        while(++level < d) {
            int size = queue.size();
            while(size-- > 0) {
                TreeNode cur = queue.poll();
                if(cur.left != null) queue.add(cur.left);
                if(cur.right != null) queue.add(cur.right);
            }
        }
        // 变换这一层以及下一层
        while(!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            TreeNode temp_left = cur.left;
            TreeNode temp_right = cur.right;
            cur.left = new TreeNode(v);
            cur.left.left = temp_left;
            cur.right = new TreeNode(v);
            cur.right.right = temp_right;
        }
        return root;
    }
}