LeetCode--populating-next-right-pointers-in-each-node

题目描述

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 toNULL.

Initially, all next pointers are set toNULL.

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

/**
 * 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) {}
 * };
 */

//不能用递归，否则递归栈占用空间，空间复杂度就不为常数了
class Solution {
public:
    void connect(TreeLinkNode *root) {
        if (!root)
            return;
        TreeLinkNode *p = root, *q;
        while (p->left) {
            q = p;
            while (q) {   //某一层的连接指针
                q->left->next = q->right;
                if (q->next)
                    q->right->next = q->next->left;
                q = q->next;
            }
            p = p->left;   //切换到下一层
        }
    }
};