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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)?

Solution

using LeetCodeNet.Com_github_leetcode;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public void Flatten(TreeNode root) {
        if (root != null) {
            FindTail(root);
        }
    }

    private TreeNode FindTail(TreeNode root) {
        TreeNode left = root.left;
        TreeNode right = root.right;
        TreeNode tail;
        // find the tail of left subtree, tail means the most left leaf
        if (left != null) {
            tail = FindTail(left);
            // stitch the right subtree below the tail
            root.left = null;
            root.right = left;
            tail.right = right;
        } else {
            tail = root;
        }
        // find tail of the right subtree
        if (tail.right == null) {
            return tail;
        } else {
            return FindTail(tail.right);
        }
    }
}

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