Flatten Binary Tree to Linked List

Posted 2020-06-17

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

For example,
Given

         1
        /        2   5
      / \        3   4   6

 

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;
    }
}