114

114. Flatten Binary Tree to Linked List (Medium)

Given the root of a binary tree, flatten the tree into a "linked list":

• The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
• The "linked list" should be in the same order as a pre-order traversal of the binary tree.

 

Example 1:

Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [0]
Output: [0]

 

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -100 <= Node.val <= 100

 

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

Companies:
Facebook, Microsoft, Bloomberg, Amazon

Related Topics:
Linked List, Stack, Tree, Depth-First Search, Binary Tree

Similar Questions:

• Flatten a Multilevel Doubly Linked List (Medium)
• Correct a Binary Tree (Medium)

Solution 1. DFS

// OJ: https://leetcode.com/problems/flatten-binary-tree-to-linked-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
    TreeNode* dfs(TreeNode* root) { // returns a pointer to the last node after flattening
        if (!root || (!root->left && !root->right)) return root;
        auto leftLast = dfs(root->left), rightLast = dfs(root->right), left = root->left, right = root->right;
        root->left = nullptr;
        root->right = left ? left : right;
        if (left) leftLast->right = right;
        return right ? rightLast : leftLast;
    }
public:
    void flatten(TreeNode* root) {
        dfs(root);
    }
};

Solution 2. Stack

// OJ: https://leetcode.com/problems/flatten-binary-tree-to-linked-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
    void flatten(TreeNode* root) {
        if (!root) return;
        stack<TreeNode*> s{{root}};
        TreeNode *prev = nullptr;
        while (s.size()) {
            auto node = s.top();
            s.pop();
            if (prev) prev->right = node;
            prev = node;
            if (node->right) s.push(node->right);
            if (node->left) s.push(node->left);
            node->left = node->right = nullptr;
        }
    }
};

Solution 3. Left-to-right Pre-order Traversal

// OJ: https://leetcode.com/problems/flatten-binary-tree-to-linked-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
    TreeNode *prev = nullptr;
public:
    void flatten(TreeNode* root) {
        if (!root) return;
        auto right = root->right;
        if (prev) prev->right = root;
        prev = root;
        flatten(root->left);
        root->left = nullptr;
        flatten(right);
    }
};

Solution 4. Right-to-left Post-order Traversal

Note: Left-to-right pre-order traversal has the reverse result of right-to-left post-order traversal.

// OJ: https://leetcode.com/problems/flatten-binary-tree-to-linked-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
    TreeNode *prev = nullptr;
public:
    void flatten(TreeNode* root) {
        if (!root) return;
        flatten(root->right);
        flatten(root->left);
        root->right = prev;
        root->left = nullptr;
        prev = root;
    }
};