LeetCode OJ 117. Populating Next Right Pointers in Each Node II

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.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
Example:

Given the following binary tree,

     1
   /  \
  2    3
 / \    \
4   5    7

After calling your function, the tree should look like:

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

解答

这题就是层序遍历二叉树。。。又是一遍就AC了。

下面是AC的代码：

/**
 * 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:
    queue<TreeLinkNode *> q;
    void connect(TreeLinkNode *root) {
        if(root == NULL){
            return ;
        }
        q.push(root);
        int i;
        while(i = q.size()){
            while(i--){
                TreeLinkNode *temp = q.front();
                q.pop();
                if(i == 0){
                    temp->next = NULL;
                }
                else{
                    temp->next = q.front();
                }
                if(temp->left != NULL){
                    q.push(temp->left);
                }
                if(temp->right != NULL){
                    q.push(temp->right);
                }
            }
        }
    }
};

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