LeetCode 623 在二叉树中增加一行[BFS DFS] HERODING的LeetCode之路

解题思路：
首先是BFS的思路，将root放入队列中，逐层遍历，当下一层就是目标层时，取出当前层的所有元素，并加上新的左右节点，同时与下一层相连，代码如下：

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    TreeNode* addOneRow(TreeNode* root, int val, int depth) {
    
        if(depth == 1) {
    
            TreeNode *node = new TreeNode(val);
            node->left = root;
            return node;
        }
        queue<TreeNode*> q;
        q.emplace(root);
        int cur = 1;
        while(!q.empty()) {
    
            int n = q.size();
            for(int i = 0; i < n; i ++) {
    
                TreeNode* temp = q.front();
                if(cur + 1== depth) {
    
                    TreeNode* l = new TreeNode(val);
                    TreeNode* r = new TreeNode(val);
                    l->left = temp->left;
                    r->right = temp->right;
                    temp->left = l;
                    temp->right = r;
                } else {
    
                    if(temp->left != nullptr) {
    
                        q.emplace(temp->left);
                    }
                    if(temp->right != nullptr) {
    
                        q.emplace(temp->right);
                    }
                }
                q.pop();
            }
            if(cur + 1 == depth) {
    
                break;
            }
            cur ++;
        }
        return root;
    }
};

BFS的思路实现起来略显繁琐，相比之下，DFS就简洁的多，但是理解稍微有些困难，对于每次处理情况，只在于当左右两节点，如果满足深度，那么就创建新节点，然后连接左节点或者右节点，是左是右可以借助深度来判断，深度为1则左，0位右，这样可以减少大量不必要操作，代码如下：

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    TreeNode* addOneRow(TreeNode* root, int val, int depth) {
    
        if(depth == 0 || depth == 1) {
    
            TreeNode* node = new TreeNode(val);
            if(depth == 1) node->left = root;
            else node->right = root;
            return node;
        }
        if(root != nullptr && depth > 1) {
    
            root->left = addOneRow(root->left, val, depth > 2 ? depth - 1 : 1);
            root->right = addOneRow(root->right, val, depth > 2 ? depth - 1 : 0);
        }
        return root;
    }
};