• 思路：
  • 根据二叉树前序遍历：根-左子树-右子树；
  • 要按照前序遍历将二叉树展开，则遍历节点右子树需要挂载到左子树“最右”节点右子树上；
  • 则当前节点 current 左子树 next = current->left 的最右节点 rightmost ：

    • TreeNode* rightmost = next;
      while (rightmost->right != nullptr) {rightmost = rightmost->right;
      }

  • 将当前节点右子树挂载到左子树“最右”节点的右子树上：rightmost->right = current->right;
  • 则当前节点 current 展开完成，将其左子树按照要求置 nullptr，右子树挂载其左子树节点：current->left = nullptr;current->right = next;
  • 迭代下一个需要展开的节点对应的树；

/*** 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 *current = root;while (current != nullptr) {if (current->left != nullptr) {TreeNode* next = current->left;TreeNode* rightmost = next;while (rightmost->right != nullptr) {rightmost = rightmost->right;}rightmost->right = current->right;current->left = nullptr;current->right = next;}current = current->right;}}
};