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Flatten Binary Tree to Linked List

时间:2017-02-01 22:37:27      阅读:156      评论:0      收藏:0      [点我收藏+]

标签:example   pre   process   code   show   题目   遍历   null   lis   

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

分析: 这道题目的求解思路是对于某个既含有左子树又含有右子树的节点T,寻找左子树的中最靠右的叶子节点,将T的右子树挂在该叶子节点的右节点上,再将T的左边节点移动到右边,依次遍历右子树

/**
 * 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) {
        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;
            
        }   
        
        
    }
};

 

Flatten Binary Tree to Linked List

标签:example   pre   process   code   show   题目   遍历   null   lis   

原文地址:http://www.cnblogs.com/willwu/p/6360434.html

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