Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.

Example 1:
Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]

       5
      / \
    3    6
   / \    \
  2   4    8
 /        / \ 
1        7   9

Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

 1
  \
   2
    \
     3
      \
       4
        \
         5
          \
           6
            \
             7
              \
               8
                \
                 9  
 

Note:

1. The number of nodes in the given tree will be between 1 and 100.
2. Each node will have a unique integer value from 0 to 1000.

題解：好坑啊。。。一直以為輸入是二叉排序樹。。。題目也不講清楚

dfs瘋狂WA

class Solution {
public:
    TreeNode* increasingBST(TreeNode* root) {
        TreeNode *p = root;
        vector<int> v;
        if (p == NULL) {
          return NULL;
        }
        else {
          dfs(p, v);
        }
        for (int i = 0; i < v.size(); i++) {
          cout << v[i] << " ";
        }
        sort(v.begin(), v.end());
        TreeNode *q = new TreeNode(v[0]);
        TreeNode *ans = q;
        for (int i = 1; i < v.size(); i++) {
          q->right = new TreeNode(v[i]);
          q = q->right;
        }
        return ans;
    }
    void dfs(TreeNode *p, vector<int> &v) {
      if (p != NULL) {
        v.push_back(p->val);
        dfs(p->left, v);
        dfs(p->right, v);
      }
    }
};
 

中序遍歷ok了

class Solution {
public:
    TreeNode* increasingBST(TreeNode* root) {
        TreeNode *p = root;
        vector<int> v;
        if (p == NULL) {
          return NULL;
        }
        else {
          inorder(p, v);
        }
        TreeNode *q = new TreeNode(v[0]);
        TreeNode *ans = q;
        for (int i = 1; i < v.size(); i++) {
          q->right = new TreeNode(v[i]);
          q = q->right;
        }
        return ans;
    }
    void inorder(TreeNode *p, vector<int> &v) {
      if (p != NULL) {
        inorder(p->left, v);
        v.push_back(p->val);
        inorder(p->right, v);
      }
    }
};