[数据结构与算法]Leetcode 二叉树展开为链表

非递归的方法：

若 root 左子树非空，则找到左子树的最右节点，将右子树接到右节点
再将左子树移到根节点的右节点上，root 不断更新成右节点，最后得到输出

/**
 * 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:
    void flatten(TreeNode* root) {
        TreeNode* last = nullptr;
        while(root){
            if(root->left) {
                last = root->left;
                while(last->right) last = last->right;
                last->right = root->right;
                root->right = root->left;
                root->left = nullptr;
            }
            root = root->right;
        }
    }
};

递归 + 前序遍历 dfs：

/**
 * 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 {
    TreeNode* dfs(TreeNode* tree) {
        if(!tree) return nullptr;
        TreeNode* l_tree = tree->left;
        TreeNode* r_tree = tree->right;

        tree->left = nullptr;
        tree->right = dfs(l_tree);
        TreeNode* current = tree;
        while(current->right) current = current->right;
        current->right = dfs(r_tree);
        return tree;
    }
public:
    void flatten(TreeNode* root) {
        dfs(root);
    }
};