Populating Next Right Pointers in Each Node

本文介绍如何在完美二叉树中填充每个节点的Next指针,使其指向右侧相邻节点。采用常数额外空间复杂度的方法,通过逐层遍历的方式设置Next指针。

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,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

思路:要充分利用题目给的条件。  有 Next  指针。 要一层一层的访问, 首先,root .next = null .  其次,在当前层, 把下一层的next 全设置了。

易错点: 1, 要判断cur 节点是否有left , 是否有下一层, right 也要在这个条件下。  2  不要忘记每一层的最右 if(cur.next != null){                        cur.right.next = cur.next.left;
                    }else{
                        cur.right.next = null;
                    }
3 , 注意每一层的头, 最左节点。

/**
 * 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 curHead = root;
        TreeLinkNode cur = curHead;
        root.next = null;
        
        while(true){
            while(cur != null){
                if(cur.left != null){
                    cur.left.next = cur.right;
                    if(cur.next != null){
                        cur.right.next = cur.next.left;
                    }else{
                        cur.right.next = null;
                    }
                }
                cur = cur.next;
            }
            if(curHead.left != null){
                curHead = curHead.left;
                cur = curHead;
            }else{
                break;
            }
        }
    }
}



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