
迭代法一:额外容器,前序遍历暴力求解(空间O(n))
/**
* 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) {
stack<TreeNode*> stk;
vector<TreeNode*> v;
TreeNode* tmp = nullptr;
TreeNode* cur = root;
while(cur || stk.size())
{
while(cur)
{
v.push_back(cur);
stk.push(cur->right);
cur = cur->left;
}
cur &