LeetCode 114 Flatten Binary Tree to Linked List (DFS 分治)

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/Tc_To_Top/article/details/89646962

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

For example, given the following tree:

    1
   / \
  2   5
 / \   \
3   4   6

The flattened tree should look like:

1
 \
  2
   \
    3
     \
      4
       \
        5
         \
          6

题目链接：https://leetcode.com/problems/flatten-binary-tree-to-linked-list/

题目分析：flatten左子树，接到右边，不断递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public void flatten(TreeNode root) {
        if (root == null || (root.left == null && root.right == null)) {
            return;
        }
        while (root != null) {
            TreeNode lTree = root.left;
            TreeNode rTree = root.right;
            flatten(lTree);
            // link flattened left node to root's right
            root.right = lTree;
            root.left = null;
            // go to the previous right node
            while (root.right != null) {
                root = root.right;
            }
            root.right = rTree;
            root = root.right;
        }
    }
}