LeetCode-Populating Next Right Pointers in Each Node II

最新推荐文章于 2025-09-04 16:45:21 发布

转载最新推荐文章于 2025-09-04 16:45:21 发布 · 66 阅读

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原文链接：http://www.cnblogs.com/superzrx/p/3326813.html

文章标签：

#c/c++

本文探讨了如何解决给定二叉树中节点之间的连接问题，并提供了在任意二叉树上的解决方案，同时强调了使用常数额外空间的重要性。

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,

After calling your function, the tree should look like:

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

/**
 * 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:
    int Count(TreeLinkNode *root){
        if(root==NULL)return 0;
        else return Count(root->left)+Count(root->right)+1;
    }
    void connect(TreeLinkNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root==NULL)return;
        int count=Count(root);
        vector<TreeLinkNode*> queue;
        queue.resize(count);
        int head=0,tail=1;
        queue[0]=root;
        int current=1;
        int next=0;
        while(head<count){
            
            if(queue[head]->left!=NULL){
                queue[tail]=queue[head]->left;
                tail++;
                next++;
            }
             if(queue[head]->right!=NULL){
                queue[tail]=queue[head]->right;
                tail++;
                next++;
            }
            current--;
            if(current==0){
                current=next;
                next=0;
            }
            else{
                queue[head]->next=queue[head+1];
            }
            head++;
        }
    }
};

转载于:https://www.cnblogs.com/superzrx/p/3326813.html