Leetcode: Flatten Binary Tree to Linked List

Q: 如何返回一颗子树的最后一个节点?

A: 如果有有右子树, 则最右节点在左子树上, 否则在左子树上

 

思路:

1. 题目要求转换要 in-place

2. 函数 preorder(root) 返回以 root 为根子树的最右节点, 同时完成 flatten 工作

 

代码:

代码的逻辑有些乱

class Solution {
public:
    void flatten(TreeNode *root) {
		if(root == NULL)
			return;

        preorder(root);
    }
	TreeNode* preorder(TreeNode *root) {
		if(root->left == NULL && root->right == NULL)
			return root;

		TreeNode *rc = root->right;
		TreeNode *rl;
		if(root->left != NULL) {
			root->right = root->left;
			rl = preorder(root->left);
			rl->right =rc;
		}
		root->left = NULL;
		if(rc != NULL)
			return preorder(rc);
		else
			return rl; 
	}
};