623. Add One Row to Tree(Difficulty: Medium)

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 asN'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, itsoriginal 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 = 1d = 2Output:        4      / \     1   1    /     \   2       6  / \     /  3   1   5   

Example 2:

Input: A binary tree as following:      4     /       2       / \     3   1    v = 1d = 3Output:       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.

思路：看到树要首先想到递归，所以这个题目我也是采用递归的方法，所以先写特殊的情况，（1）当d等于1时，是把v当做根，而原来的树是新树的左子根。（2）当d等于2时，根的左右子树都变了，分别以v创建两个新树，一个是根的左子树，一个是根的右子树。这道题还是挺简单的，原来是左子树还是左子树，原来是右子树还是右子树，不需要与v作比较，这样就少了好几种情况。

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* addOneRow(TreeNode* root, int v, int d) {
        if(d==1)
        {
            TreeNode * newtree = new TreeNode(v);
            newtree->left = root;
            return newtree;
        }
        else if (d == 2) {
            TreeNode * newtreeleft = new TreeNode(v);
            TreeNode * newtreeright = new TreeNode(v);
            newtreeleft->left = root->left;
            root->left = newtreeleft;
            newtreeright->right = root->right;
            root->right = newtreeright;
            return root;
        }
        else if (d > 2) {
            root->left = addOneRow(root->left, v, d - 1);
            root->right = addOneRow(root->right, v, d - 1);
            return root;
        }
        
    }
};