623. Add One Row to Tree

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623. Add One Row to Tree

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

思路
BFS 或者 DFS到目的层次的上一层，方便在目标层插入结点。

代码

1、DFS的解法

 public TreeNode addOneRow(TreeNode root, int v, int d) {
        if (d == 1) {
            TreeNode newRoot = new TreeNode(v);
            newRoot.left = root;
            return newRoot;
        }
        helper(root, 1, v, d);
        return root;
    }
    
    private void helper(TreeNode node, int depth, int v, int d) {
        if (node == null) {
            return;
        }
        if (depth == d - 1) {
            TreeNode left = node.left;
            TreeNode right = node.right;
            
            TreeNode newNodeL = new TreeNode(v);
            TreeNode newNodeR = new TreeNode(v);
            node.left = newNodeL;
            node.right = newNodeR;
            
            newNodeL.left = left;
            newNodeR.right = right;
            return;
        }
        
        helper(node.left, depth + 1, v, d);
        helper(node.right, depth + 1, v, d);
    }

2、BFS解法

   public TreeNode addOneRow(TreeNode root, int v, int d) {
        if(d == 1){
            TreeNode cur = new TreeNode((v));
            if(root == null )return cur;  
            
            cur.left = root;
            return cur;
                
        }
        Queue<TreeNode> q = new LinkedList<>();
        q.add(root);
        while(!q.isEmpty()){
            int size = q.size();
            d--;
            for(int i = 0; i <size; i++){
                TreeNode p = q.poll();
                TreeNode left = p.left, right = p.right;
                if(d  == 1){
                   
                   
                         p.left  = new TreeNode(v);
                         p.left.left = left;
                    
                         p.right = new TreeNode(v);
                         p.right.right = right;
                        
                }
      
                else{
                    if(left != null) q.add(left);
                    if(right != null) q.add(right);
                }
            }
            if(d == 1) break;
        }
        return root;
    }