leetcode 623. Add One Row to Tree二叉树添加指定深度元素+ 深度优先遍历DFS

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.

本题题意很简单，就是添加一层结点，对于d=1的时候题目没有说明，那么直接把root添加到新节点的left孩子即可，别的直接递归解决

我想了一下没想到很棒的递归做法，网上看到了一个做法，十分棒得递归

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>

using namespace std;

/*
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
*/

class Solution
{
public:
vector<TreeNode*> res;
TreeNode* addOneRow(TreeNode* root, int v, int d)
{
if (d == 1)
{
TreeNode* newRoot = new TreeNode(v);
newRoot->left = root;
newRoot->right = NULL;
return newRoot;
}
else if (d == 2)
{
TreeNode* newleft = new TreeNode(v);
newleft->left = root->left;
newleft->right = NULL;
root->left = newleft;

TreeNode* newright = new TreeNode(v);
newright->left = NULL;
newright->right = root->right;
root->right = newright;

return root;
}
else
{
if (root->left != NULL)
addOneRow(root->left, v, d - 1);

if (root->right != NULL)
addOneRow(root->right, v, d - 1);

return root;
}
}
};