[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;
    }
}