Flatten Binary Tree to Linked List

最新推荐文章于 2025-09-10 15:24:29 发布

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最新推荐文章于 2025-09-10 15:24:29 发布

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CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/sdlwlxf/p/5224557.html

本文介绍了一种将二叉树结构展平为链表的方法，通过两种不同的算法实现：一种使用栈来辅助完成前序遍历，另一种则是在不使用额外栈空间的情况下直接操作节点指针进行展平。

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

For example,
Given

The flattened tree should look like:

   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6


利用前序遍历解决的方法实现

void flatten(TreeNode* root) {
    if (root == nullptr)
        return;
    stack<TreeNode*> sta;
    sta.push(root);
    TreeNode* lastRoot = root;
    while (!sta.empty())
    {
        root = sta.top();
        if (lastRoot != root->right)
        {
            if (lastRoot != root->left) {
                if (root->left != nullptr) {
                    sta.push(root->left);
                    continue;
                }
            }
            else
            {
                TreeNode* tempNode = root->right;
                root->right = lastRoot;
                root->left = nullptr;
                TreeNode* endNode = lastRoot;
                while (endNode->right != nullptr)
                    endNode = endNode->right;
                endNode->right = tempNode;
            }
            if (root->right != nullptr) {
                sta.push(root->right);
                continue;
            }
        }
        lastRoot = root;
        sta.pop();
    }
}

不使用遍历的方法，即不使用栈解决的方法

void flatten(TreeNode *root) {
    TreeNode*now = root;
    while (now)
    {
        if (now->left)
        {
            //Find current node's prenode that links to current node's right subtree
            TreeNode* pre = now->left;
            while (pre->right)
            {
                pre = pre->right;
            }
            pre->right = now->right;
            //Use current node's left subtree to replace its right subtree(original right 
            //subtree is already linked by current node's prenode
            now->right = now->left;
            now->left = NULL;
        }
        now = now->right;
    }
}