[LeetCode] Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

         1
        / 
       2   5
      /    
     3   4   6

The flattened tree should look like:

   1
    
     2
      
       3
        
         4
          
           5
            
             6

click to show hints.

Hints:

If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

思路一:递归的思想。使用后序递归的方法。先将左右子树转换为链表，再将左右子树连接

　　　  时间复杂度O(n),空间复杂度O(logN)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void flatten(TreeNode *root) {
        if (root == nullptr) return;
        
        flatten(root->left);
        flatten(root->right);
        
        //三方合并，将左子树形成的链表插入到root和root->right之间
        TreeNode *p = root->right;
        while (p->right) p = p->right;
        p->right = root->right;
        root->right = root->left;
        root->left = nullptr;
    }
};

思路二：迭代的方法。依据题目与前序遍历之间的关系。使用前序遍历的方法。

　　　　

void flatten(TreeNode *root) {
        if(root == NULL) return;
    while(root){
        if(root->left){
            TreeNode *pre = root->left;
            while(pre->right)
                pre = pre->right;
            pre->right = root->right;
            root->right = root->left;
            root->left = NULL;
        }
        root = root->right;
    }
    }