【Leetcode】Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

The flattened tree should look like:

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     void flatten(TreeNode *root) {
13         if (root == nullptr) {
14             return;
15         }
16         if (root->left != nullptr) {
17             flatten(root->left);
18         }
19         if (root->right != nullptr) {
20             flatten(root->right);
21         }
22         if (root->left != nullptr) {
23             TreeNode *tail = root->left;
24             while (tail->right != nullptr) {
25                 tail = tail->right;
26             }
27             tail->right = root->right;
28             root->right = root->left;
29             root->left = nullptr;
30         }
31     }
32 };

View Code

flatten后的链表顺序与树的先序遍历一致。树的问题通常可以考虑递归。

也可以借助栈迭代求解：

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     void flatten(TreeNode *root) {
13         if(root == nullptr) {
14             return;
15         }
16         stack<TreeNode *> s;
17         s.push(root);
18         while (!s.empty()) {
19             TreeNode *p = s.top();
20             s.pop();
21             if (p->right != nullptr) {
22                 s.push(p->right);
23             }
24             if (p->left != nullptr) {
25                 s.push(p->left);
26             }
27             if (!s.empty()) {
28                 p->right = s.top();
29             }
30             p->left = nullptr;
31         }
32     }
33 };

View Code