看图就知道想要做什么事了。
Given a binary tree
struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}
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
Note:
/** * 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: //每一层把left的right的next设置为right的left void lr2rl(TreeLinkNode *root) { if (!root) return; TreeLinkNode *lr = root -> left, *rl = root -> right; while(lr && rl) { lr -> next = rl; lr = lr -> right; rl = rl -> left; } lr2rl(root -> left); lr2rl(root -> right); } void connect(TreeLinkNode *root) { lr2rl(root); } };
leetcode Populating Next Right Pointers in Each Node
原文：http://www.cnblogs.com/higerzhang/p/4134025.html