【Leetcode】114. Flatten Binary Tree to Linked List

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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;
            }
        }
        }
        
    }
};