623. Add One Row to Tree

题目描述【Leetcode】

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:

Example 2:

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.

这道题是在某一个深度插入所给的节点，并不难，但是需要考虑d为1的情况：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
void f(TreeNode* root,int h, int v, int d){
    if(!root) return;
    if(h == d-1){
         TreeNode * lt = new TreeNode(v);
         TreeNode* rt = new TreeNode(v);
         lt->left = root->left;
         rt->right = root->right;
         root->left = lt;
         root->right = rt;
         return ;
    }
        f(root->left,h+1,v,d);
        f(root->right,h+1,v,d);

}

class Solution {
public:
    TreeNode* addOneRow(TreeNode* root, int v, int d) {
        if(d == 1){
            TreeNode * tr = new TreeNode(v);
            tr->left = root;
            return tr;
        }
        f(root,1,v,d);
        return root;
    }
};