[LeetCode] Flatten Binary Tree to Linked List

发布时间：2020-12-14 04:50:06 所属栏目：大数据来源：网络整理

导读：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 将一个二叉树重组成一个链表题目要求按照二叉树的先序遍历的顺序重组二叉树思路

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:

将一个二叉树重组成一个链表

题目要求按照二叉树的先序遍历的顺序重组二叉树

思路1：

找到最左侧结点，将其父结点与父结点右结点断开，将其连接至父结点右侧，变成父结点的右结点，然后把原右结点插入到新右结点的右侧。递归这一操作即可

/**
 * 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：

从根结点出发，判断其左结点是否存在，如果存在，则将根结点与其右结点断开，根的左结点变成其右结点。在将右结点链接至左结点最右边的右结点处。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x),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;
        }
    }
};

（编辑：李大同）

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