[LeetCode] Flatten Binary Tree to Linked List 解题报告

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.

» Solve this problem

 

对于这个问题，网络上有很多递归和迭代的方法，本文介绍一种非递归和迭代的方法，主要的思路很类似树的线索化。

　　 1                  1                  1
   /  \                 \                  \
  2    5     转换为       2         转换为    2
 / \    \               / \                  \
3   4    6             3   4                  3
                            \                  \
                             5                  4
                              \                  \
                               6                  5
                                                   \
                                                    6

 

具体思路：

将节点的左子树的最右节点的右孩子指针（初始为NULL）指向节点的右孩子，依次遍历下即可。

参考代码如下：

/**
 * 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) {
        TreeNode *cur, *pre;
        cur = root;
        while(cur) {
            if (cur->left) {
                pre = cur->left;
                while(pre->right != NULL) {
                    pre = pre->right;
                }
                pre->right = cur->right;
                cur->right = cur->left;
                cur->left = NULL;
            }
            else {
                cur = cur->right;
            }
        }
    }
};