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
将二叉树的左子树移到右子树前面,直到最后成为一颗只有右子树的树。
利用先序遍历,返回为这个子树变化后的最后一个节点。
/**
* 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:
TreeNode* dfs(TreeNode* root)
{
if(!root) return NULL;
if(!root->left && !root->right) return root;
TreeNode* q=dfs(root->right);
TreeNode* p;
if(root->left)
{
p=dfs(root->left);
p->right=root->right;
root->right=root->left;
root->left=NULL;
}
if(q) return q;
return p;
}
void flatten(TreeNode* root) {
dfs(root);
}
};