Given a binary tree 

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

After calling your function, the tree should look like:

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

思路

这也是一道非常典型的利用层次遍历解决树的问题的题目。我们可以利用队列进行BFS的操作，从而求解。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
/**
 * 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) {
        queue<TreeLinkNode *> layer;
        if(root != NULL){
            layer.push(root);
            layer.push(NULL);
        }
        TreeLinkNode *prev = NULL;
        while(!layer.empty()){
            TreeLinkNode *cur = layer.front();
            layer.pop();
            if(prev != NULL){
                prev->next = cur;
            }
            prev = cur;
            if(cur == NULL){
                if(!layer.empty()) layer.push(NULL);
            }
            else{
                if(cur->left != NULL) layer.push(cur->left);
                if(cur->right != NULL) layer.push(cur->right);
            }
        }
    }
};