[leetcode]114. Flatten Binary Tree to Linked List将二叉树展成一个链表

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

将左子树所形成的链表插入到root和root->right之间

思路

      1
   /
  2(root)              假设当前root为2
 /   \
3(p)  4     

      1
   /
  2(root)
 /     \
3(p)—— 4(root.right)    p.right = root.right

     1
    /
    2(root)
 /    \
3(root.right)- 4      root.right = root.left

      1
    /
    2(root)
      \
3(root.right)- 4      root.left - null

代码

 class Solution {
     public void flatten(TreeNode root) {
         if (root == null) return;  // 终止条件
         // recursion
         flatten(root.left);
         flatten(root.right);

         if (root.left == null) return;

         // 三方合并，将左子树所形成的链表插入到root和root->right之间
         TreeNode p = root.left;
         while(p.right != null) {
             p = p.right; //寻找左链表最后一个节点
         }
         p.right = root.right;
         root.right = root.left;
         root.left = null;
     }
 }