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:
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
思路:前序的递归思路。时间复杂度O(n),,空间复杂度O(1)
1 /** 2 * Definition for binary tree with next pointer. 3 * struct TreeLinkNode { 4 * int val; 5 * TreeLinkNode *left, *right, *next; 6 * TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 void connect(TreeLinkNode *root, TreeLinkNode *sibling) { 12 //处理当前节点 13 if (root == NULL) 14 return; 15 else 16 root->next = sibling; 17 18 //处理左子节点 19 connect(root->left, root->right); 20 21 //处理右子节点 22 if (sibling) { 23 connect(root->right, sibling->left); 24 } else { 25 connect(root->right, NULL); 26 } 27 } 28 void connect(TreeLinkNode *root) { 29 connect(root, NULL); 30 } 31 };
[LeetCode] Populating Next Right Pointers in Each Node
原文：http://www.cnblogs.com/vincently/p/4231719.html