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

For example, given the following tree:

1/ \2 5/ \ \ 3 4 6

The flattened tree should look like:

1\2\3\4\5\6

將一個二叉樹重組成一個鏈表

題目要求按照二叉樹的先序遍歷的順序重組二叉樹

思路1：

找到最左側結點，將其父結點與父結點右結點斷開，將其連接至父結點右側，變成父結點的右結點，然后把原右結點插入到新右結點的右側。遞歸這一操作即可

/*** 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) {if (!root)return;if (root->left)flatten(root->left);if (root->right)flatten(root->right);TreeNode* temp = root->right;root->right = root->left;root->left = NULL;while (root->right)root = root->right;root->right = temp;} };

思路2：

從根結點出發，判斷其左結點是否存在，如果存在，則將根結點與其右結點斷開，根的左結點變成其右結點。在將右結點鏈接至左結點最右邊的右結點處。

/*** 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) {if (!root)return;TreeNode* curr = root;while (curr){if (curr->left){TreeNode* temp = curr->left;while (temp->right)temp = temp->right;temp->right = curr->right;curr->right = curr->left;curr->left = NULL;}curr = curr->right;}} };

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轉載于:https://www.cnblogs.com/immjc/p/9076162.html

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