117. Populating Next Right Pointers in Each Node II

最新推荐文章于 2021-01-29 16:34:55 发布

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最新推荐文章于 2021-01-29 16:34:55 发布

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分类专栏： leetCode 文章标签： tree

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本文介绍了一种在任意二叉树中填充每个节点的下一个指针的方法，确保同一层的相邻节点通过next指针连接。此算法只使用常数额外空间，并通过迭代而非递归的方式实现。

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/*
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

*/


/**
 * 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) return;
        TreeLinkNode *pre=NULL;//下一行中当前节点的前一个节点
        TreeLinkNode *headL=NULL;//下一行的从左边第一个非空节点
        TreeLinkNode *cur=root;
        while(cur)
        {
            while(cur)
            {
                if(cur->left){
                    if(pre==NULL){
                        headL=cur->left;
                    }else{
                        pre->next=cur->left;
                    }
                    pre=cur->left;
                }
                if(cur->right){
                    if(pre==NULL){
                        headL=cur->right;
                    }else{
                        pre->next=cur->right;
                    }
                    pre=cur->right;
                }
                cur=cur->next;
            }
            cur=headL;
            headL=NULL;
            pre=NULL;
        }
    }
};