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Given a binary tree, flatten it to a linked list in-place.
For example, given the following tree:
1 / \ 2 5 / \ \ 3 4 6
The flattened tree should look like:
1 \ 2 \ 3 \ 4 \ 5 \ 6
题目大意:
按照先根节点,左节点,右节点的顺序排列出一个仅包含右节点的树。
解题思路:
vector先保存父节点,再左节点,右节点保存的顺序遍历。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
vector<int> nums;
void helper(TreeNode *root){
if(root->left==NULL&&root->right==NULL){
nums.push_back(root->val);
return ;
}
nums.push_back(root->val);
if(root->left) helper(root->left);
if(root->right) helper(root->right);
}
public:
void flatten(TreeNode* root) {
if(root){
helper(root);
// sort(nums.begin(), nums.end());
root->val = nums[0];
TreeNode* tmp=root;
// root = tmp;
for(int i=1;i<nums.size();i++){
if(tmp->left){
tmp->left=nullptr;
}
if(tmp->right==NULL){
TreeNode *t = new TreeNode(nums[i]);
tmp->right = t;
tmp = tmp->right;
}else{
tmp->right->val = nums[i];
tmp = tmp->right;
}
}
}
}
};