Populating Next Right Pointers in Each Node

最新推荐文章于 2022-02-28 18:16:11 发布

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最新推荐文章于 2022-02-28 18:16:11 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/sunyujia19/article/details/13763775

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本文介绍了一种解决二叉树节点连接问题的算法实现，通过遍历每一层的节点并将其相邻的节点进行连接，使得每个节点都可以直接访问到其下一个节点。该算法采用两种不同的实现方式，有效地完成了节点间的连接。

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连相同列，再遍历下一层的相同列

/**
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if (!root) return;
        if (root->left) {
            TreeLinkNode *l=root->left, *r=root->right;
            while (l) {
                l->next=r;
                l=l->right;
                r=r->left;
            }
            connect(root->left);
            connect(root->right);
        }
    }
};

II:

/**
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        TreeLinkNode *cur = root;
        while (cur) {
            TreeLinkNode *pre=NULL, *next=NULL;
            for ( ; cur!=NULL; cur=cur->next) {
                if (!next) next=cur->left?cur->left:cur->right;
                if (cur->left) {
                    if (pre) pre->next=cur->left;
                    pre=cur->left;
                }
                if (cur->right) {
                    if (pre) pre->next=cur->right;
                    pre=cur->right;
                }
            }
            cur=next;
        }
    }
    
};