Flatten Binary Tree to Linked List

一、问题描述

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

二、思路

本题显而易见，将二叉树转变为只有右子树的二叉树。

当左子树的右子树不为空，将根节点的右子树赋值给最后一个右子树为空的节点的右子树。在循环中判断根节点的左子树如果不为空，将右子树指向当前节点的左子树，同时修改当前节点的左子树，当前节点指向右子树，循环执行，直到当前节点为叶子节点。

三、代码

/**
* 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;
}
}
};