116. Populating Next Right Pointers in Each Node

问题描述：

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 to NULL.

Initially, all next pointers are set to NULL.

Note:

You may only use constant extra space.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

Example:

Given the following perfect binary tree,

After calling your function, the tree should look like:

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

解题思路：

这道题当然首先考虑层序遍历。

重要的一点是找到每一层的开头。

所以我采用的策略是用一个指针next指向每一层的开头，初始化时为root

当next与当前的节点为同一节点时，我们可以知道我们开始了新的一层，所以需要找到下一层的最左边的节点。

if(cur == next || next == NULL){
      next = cur->left ? cur->left : cur->right;
}

需要注意的是，这一层的最左边的节点可能不是下一层最左边节点的父节点。

代码：

/**
 * 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;
        queue<TreeLinkNode*> q;
        q.push(root);
        TreeLinkNode* next = root;
        TreeLinkNode* prev = NULL;
        while(!q.empty()){
            TreeLinkNode* cur = q.front();
            q.pop();
            if(prev) prev->next = cur;
            if(cur == next || next == NULL){
                next = cur->left ? cur->left : cur->right;
            }
            if(cur->left) q.push(cur->left);
            if(cur->right) q.push(cur->right);
            prev = cur;
            if(!q.empty() && q.front() == next) prev = NULL;
        }
    }
};