[LintCode] Flatten Binary Tree to Linked List 将二叉树展开成链表

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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文章标签：数据结构与算法

原文链接：https://yq.aliyun.com/articles/339200

本文介绍了一种将二叉树转换为链表的方法，通过使用三种不同的递归和迭代策略，使得二叉树可以按前序遍历的方式被拉平。此过程涉及到修改二叉树节点的左右子节点指向，最终实现将二叉树转化为一个类似链表的结构。

Flatten a binary tree to a fake "linked list" in pre-order traversal.

Here we use the right pointer in TreeNode as the next pointer in ListNode.

Notice

Don't forget to mark the left child of each node to null. Or you will get Time Limit Exceeded or Memory Limit Exceeded.

Example

              1
               \
     1          2
    / \          \
   2   5    =>    3
  / \   \          \
 3   4   6          4
                     \
                      5
                       \
                        6

LeetCode上的原题，请参见我之前的博客Flatten Binary Tree to Linked List。

解法一：

class Solution {
public:
    /**
     * @param root: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    void flatten(TreeNode *root) {
        if (!root) return;
        stack<TreeNode*> s;
        s.push(root);
        while (!s.empty()) {
            TreeNode *t = s.top(); s.pop();
            if (t->left) {
                TreeNode *r = t->left;
                while (r->right) r = r->right;
                r->right = t->right;
                t->right = t->left;
                t->left = NULL;
            }
            if (t->right) s.push(t->right);
        }
    }
};

解法二：

class Solution {
public:
    /**
     * @param root: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    void flatten(TreeNode *root) {
        if (!root) return;
        if (root->left) flatten(root->left);
        if (root->right) flatten(root->right);
        TreeNode *t = root->right;
        root->right = root->left;
        root->left = NULL;
        while (root->right) root = root->right;
        root->right = t;
    }
};

解法三：

class Solution {
public:
    /**
     * @param root: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    void flatten(TreeNode *root) {
        TreeNode *cur = root;
        while (cur) {
            if (cur->left) {
                TreeNode * t = cur->left;
                while (t->right) t = t->right;
                t->right = cur->right;
                cur->right = cur->left;
                cur->left = NULL;
            }
            cur = cur->right;
        }
    }
};