leetcode：Populating Next Right Pointers in Each Node（常数空间，连接二叉树每一层所有节点）【面试算法题】

题目：

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

题意把给定的一个完全二叉树每一层所有的节点连接起来，只能用常数的额外空间。

用递归的做法，把当前root的左节点和它的所有深度下的右节点 与 右节点和它的所有深度下的左节点 连接。

目的就是把因为当前root左右节点而分开的子节点全部连接起来。

画个图就很容易发现，这样做就把需要连接的节点在他们的公共祖先root的递归层中一起处理。

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(!root)return;
        if(root->left)
        {
            TreeLinkNode *l=root->left,*r=root->right;
            while(l!=NULL)
            {
                l->next=r;
                l=l->right;
                r=r->left;
            }
            connect(root->left);
            connect(root->right);
        }
    }
};
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