Leetcode 116. Populating Next Right Pointers in Each Node

Given a binary tree

struct TreeLinkNode {
  TreeLinkNode *left;
  TreeLinkNode *right;
  TreeLinkNode *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,

     1
   /  \
  2    3
 / \  / \
4  5  6  7

After calling your function, the tree should look like:

     1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \  / \
4->5->6->7 -> NULL

分析

为二叉树添加next指针，指向同高度的右侧节点。由于是满二叉树，所以只需要递归来设置左右子节点的next指针即可。

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  struct TreeLinkNode *left, *right, *next;
 * };
 *
 */
void connect(struct TreeLinkNode *root) {
    if(root==NULL||(root->left==NULL&&root->right==NULL))
    {
        return;
    }
    root->left->next=root->right;
    if(root->next==NULL)
    {
        root->right->next=NULL;
    }
    else
    {
        root->right->next=root->next->left;
    }
    connect(root->left);
    connect(root->right);
    return;
}