难度:中等
给定一个二叉树,原地将它展开为一个单链表。
例如,给定二叉树
1 / \ 2 5 / \ \ 3 4 6
将其展开为:
1 \ 2 \ 3 \ 4 \ 5 \ 6
题目分析:
直接前序遍历非递归版,然后将所有节点存储起来,最后使用右指针链接起来,左指针都置null(空间复杂度偏高)
参考代码:
/**
* 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) {
if(root == NULL)
return;
TreeNode* pMove = root;
stack<TreeNode*> node_stack;
vector<TreeNode*> node_data;
while(pMove || !node_stack.empty())
{
while(pMove)
{
node_data.push_back(pMove);
node_stack.push(pMove);
pMove = pMove->left;
}
if(!node_stack.empty())
{
pMove = node_stack.top();
node_stack.pop();
pMove = pMove->right;
}
}
for(int i = 0; i < node_data.size(); i++)
{
node_data[i]->left = nullptr;
if(i+1 < node_data.size())
node_data[i]->right = node_data[i+1];
}
}
};