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LeetCode: Populating Next Right Pointers in Each Node II [117]

2024-07-07 11:03:16   214人阅读

【题目】

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

  • You may only use constant extra space.

For example,
Given the following binary tree,

         1
       /        2    3
     / \        4   5    7

After calling your function, the tree should look like:

         1 -> NULL
       /        2 -> 3 -> NULL
     / \        4-> 5 -> 7 -> NULL

【题意】

    给定一棵二叉树，要求每层上的节点链接起来。
    与Populating Next Right Pointers in Each Node不同的是，本题输入的二叉树不一定是完整二叉树


【思路】


    与Populating Next Right Pointers in Each Node思路以及方法完全相同，代码可以直接移植求解。


【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;
            }
        }
    }
};


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