[Leetcode] 623. Add One Row to Tree 解题报告

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
4000
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：

思路还是递归，但这里特别需要注意的是递归的两个边界条件：1）d == 1，说明我们需要在当前层增加结点，并且将以当前结点为根的树都下降一层。2）当前结点为空，那么必须返回。值得注意的是我们需要首先判断条件1），因为在d == 1的时候，如果当前结点为空，我们也需要处理。如果前两个条件都没有满足，即d >= 2并且root ！= NULL，那么只需要递归调用，最后返回更新后的root即可。

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) {
return addOneRow(root, v, d, true);
}
private:
TreeNode* addOneRow(TreeNode* node, int v, int d, bool is_left) {
if (d == 1) {
TreeNode *new_node = new TreeNode(v);
if (is_left) {
new_node->left = node;
}
else {
new_node->right = node;
}
return new_node;
}
if (node == NULL) { // d >= 2, so we return if node == NULL
return NULL;
}
// node is not NULL and d >= 2, so we use recurrsion
node->left = addOneRow(node->left, v, d - 1, true);
node->right = addOneRow(node->right, v, d - 1, false);
return node;
}
};