lintcode- Flatten Binary Tree to Linked List-453

Flatten a binary tree to a fake "linked list" in pre-order traversal.

Here we use the right pointer in TreeNode as the next pointer in ListNode.

样例

              1
               \
     1          2
    / \          \
   2   5    =>    3
  / \   \          \
 3   4   6          4
                     \
                      5
                       \
                        6

注意

Don't forget to mark the left child of each node to null. Or you will get Time Limit Exceeded or Memory Limit Exceeded.

挑战

Do it in-place without any extra memory.

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {                 //采用的是先序遍历
public:
   
    TreeNode *last=nullptr;     //记录当前节点的前一个节点(指遍历过程中的前后关系)
    void flatten(TreeNode *root) {
        if(!root)
            return;
        
        if(last!=nullptr){
            last->right=root;   //调整节点关系
            last->left=nullptr;
        }    
        last=root;
        TreeNode *tmp=root->right; //防止丢失节点
        flatten(root->left);
        flatten(tmp);
    }
};