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
将一颗二叉树转化为链表，right充当next指针，元素顺序为先序遍历的循序。不使用额外的空间
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* flattenTree(TreeNode *root){
if(root->left==NULL && root->right==NULL)return root;
TreeNode*tail=root;
TreeNode*headLeft=root->left;
TreeNode*tailLeft=NULL;
TreeNode*headRight=root->right;
TreeNode*tailRight=NULL;
//【注意】左孩子清NULL
root->left=NULL;
//将左子树链表化
if(headLeft!=NULL){
tailLeft=flattenTree(headLeft);
tail->right=headLeft;
tail=tailLeft;
}
//将右子树链表化
if(headRight!=NULL){
tailRight=flattenTree(headRight);
tail->right=headRight;
tail=tailRight;
}
return tail;
}
void flatten(TreeNode *root) {
if(root==NULL)return;
flattenTree(root);
}
};LeetCode: Flatten Binary Tree to Linked List [114],布布扣,bubuko.com
LeetCode: Flatten Binary Tree to Linked List [114]
原文：http://blog.csdn.net/harryhuang1990/article/details/28587927