LeetCode题解——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

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，将左子树展成一列，将右子树展成一列，然后再拼接左子树和右子树

/**
 * 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 {
public:
    void flatten(TreeNode* root) {
        if(!root) return;
        flatten(root->left);
        flatten(root->right);
        if(root->left){
            TreeNode* t = root->left;
            while(t->right){
                t=t->right;
            }
            t->right = root->right;
            root->right = root->left;
            root->left = NULL;
        }
    }
};

思路二：不用递归，用循环。首先将根节点的左子树变为NULL，具体操作为：将根节点的左子树的最右的节点指向根节点的右子树，然后将根节点的右指针指向根节点的左子树，再将左子树置为NULL。然后处理在处理根节点的右子树，把右子节点看做根节点，重复上述步骤。

/**
 * 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 {
public:
    void flatten(TreeNode* root) {
        while(root){
            if(root->left && root->right){
                TreeNode* t = root->left;
                while(t->right){
                    t = t->right;
                }
                t->right = root->right;
            }
            
            if(root->left){
                root->right = root->left;
                root->left = NULL;
            }
            root= root->right;
        }
    }
};