116. Populating Next Right Pointers in Each Node

本文介绍了一种在完美二叉树中连接每个节点的next指针到其右侧节点的算法，实现时间复杂度O(n)，空间复杂度O(1)。通过使用两个指针start和cur，逐层遍历树的每一层，使所有同层节点通过next指针相连。

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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.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

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

用两个指针，start和cur，start标记每一层的起点（最左），cur遍历该层，逐层遍历，start从root开始

time: O(n), space: O(1)

/**
 * 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) {
        if(root == null) {
            return;
        }
        
        TreeLinkNode start = root, cur = null;
        while(start.left != null) {
            cur = start;
            while(cur != null) {
                cur.left.next = cur.right;
                if(cur.next != null) {
                    cur.right.next = cur.next.left;
                }
                cur = cur.next;
            }
            start = start.left;
        }
    }
}