LeetCode: Flatten Binary Tree to Linked List [114]

【题目】

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.

【题意】

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

【思路】

先对左子树链表化，然后将右子树链表化。然后按照:根节点->左子树链表->右子树链表 的顺序组合成完整的链表

【代码】

/**
* 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);
}
};