[leetcode] Populating Next Right Pointers in Each Node

最新推荐文章于 2021-09-24 12:05:40 发布

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本文介绍了一种在二叉树中填充每个节点的next指针的方法，使得每个节点指向其右侧相邻节点。此算法适用于完美二叉树，并且只使用常数额外空间。通过迭代方式实现，首先初始化当前层的第一个节点，然后遍历该层的所有节点，连接它们的孩子节点。

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From : https://leetcode.com/problems/populating-next-right-pointers-in-each-node/

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,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->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) {
        TreeLinkNode *cur=root, *up, *dp, *t;
        while(cur) {
            up = cur;
            dp = t = cur->left;
            while(dp) {
                dp->next = up->right;
                if(!up->next) break;
                up = up->next;
                dp = dp->next;
                dp->next = up->left;
                dp = dp->next;
            }
            cur = t;
        }
    }
};

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode up = root;
        while(up != null) {
            TreeLinkNode dp = up.left;
            TreeLinkNode newUp = dp;
            while(dp != null) {
                dp = dp.next = up.right;
                if(up.next != null) {
                    up = up.next;
                    dp = dp.next = up.left;
                } else {
                    dp = null;
                }
            }
            up = newUp;
        }
    }
}