首页 > 代码库 > LeetCode: Populating Next Right Pointers in Each Node [116]

LeetCode: Populating Next Right Pointers in Each Node [116]

2024-07-07 05:16:46   224人阅读

【题目】

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



【题意】

    给定一个二叉树，把同层的节点连接起来。
    题目保证输入的二叉树是完整二叉树


【思路】


    BFS遍历各层节点，维护两张队列，交替保存相邻各层的节点。


【代码】

/**
 * 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;
        TreeLinkNode*prev;
        TreeLinkNode*cur;
        queue<TreeLinkNode*>q1;
        queue<TreeLinkNode*>q2;
        q1.push(root);
        while(!q1.empty() || !q2.empty()){
            prev=NULL;
            if(!q1.empty()){
                while(!q1.empty()){
                    cur=q1.front(); q1.pop();
                    if(prev)prev->next=cur;
                    prev=cur;
                    //把下层节点保存到q2
                    if(cur->left)q2.push(cur->left);
                    if(cur->right)q2.push(cur->right);
                }
                cur->next=NULL;
               
            }
            else{
                while(!q2.empty()){
                    cur=q2.front(); q2.pop();
                    if(prev)prev->next=cur;
                    prev=cur;
                    //把下层节点保存到q1
                    if(cur->left)q1.push(cur->left);
                    if(cur->right)q1.push(cur->right);
                }
                cur->next=NULL;
            }
        }
    }
};


联系
我们