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.
- 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 6 7
After calling your function, the tree should look like:
1 -> NULL / 2 -> 3 -> NULL / / 4->5->6->7 -> NULL
题目要求是一个二叉树每个节点包含三个指针元素,每个节点除了左子树和右子树,还有一个指向其同层次的相邻右侧节点,
若同层次右侧没有节点,则将next设置为空,否则将next设置为右侧相邻节点。
解决思路:
一次对每个节点进行遍历,若左右子树不为空,则将左子树的next指向右子树,若该节点的next为空,则将右子树的next设置为空(看图分析)
若该节点的next不为空,则指向与该节点同层次中相邻的右侧节点的左子树(看图分析)
这里使用一个队列,将根节点先放入队列,从队列中取出一个节点node,node移除队列,node子树的next节点处理按照上面分析操作。
代码如下:
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) { 12 if(root == NULL) return; 13 queue<TreeLinkNode *> q; 14 15 root->next = NULL; 16 17 q.push(root); 18 19 while(!q.empty()){ 20 TreeLinkNode *node = q.front(); 21 q.pop(); 22 23 if(node->left != NULL && node->right != NULL){ 24 q.push(node->left); 25 q.push(node->right); 26 27 node->left->next = node->right; 28 if(node->next == NULL) 29 node->right->next = NULL; 30 else{ 31 TreeLinkNode *node_next = q.front(); 32 node->right->next = node_next->left; 33 } 34 35 } 36 37 } 38 39 } 40 };