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
思路：主要是使用层次遍历，将所有结点按照层次遍历的操作将前一个出队列的结点的next域指向队首元素，即下一个要访问的结点，这样就将所有结点连成一串了，然后将所有一直靠右的子树的next置位NULL。
C++实现代码如下：
#include<iostream> #include<new> #include<queue> using namespace std; // 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) { if(root==NULL) return; queue<TreeLinkNode*> q; q.push(root); TreeLinkNode *tmp=root; while(!q.empty()) { tmp->next=q.front(); tmp=q.front(); q.pop(); if(tmp->left) q.push(tmp->left); if(tmp->right) q.push(tmp->right); } tmp->next=NULL; while(root) { root->next=NULL; root=root->right; } } void createTree(TreeLinkNode *&root) { int i; cin>>i; if(i!=0) { root=new TreeLinkNode(i); if(root==NULL) return; createTree(root->left); createTree(root->right); } } }; int main() { Solution s; TreeLinkNode *root; s.createTree(root); s.connect(root); cout<<root->val<<endl; while(root->left) { cout<<root->left->val<<" "; TreeLinkNode *tmp=root->left->next; while(tmp) { cout<<tmp->val<<" "; tmp=tmp->next; } cout<<endl; root=root->left; } }
运行结果：
