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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本文介绍了一种在完美二叉树中填充每个节点的next指针的方法，使其指向右侧相邻节点。提供了两种解决方案：一种使用额外的空间，另一种利用递归和已填充的next指针实现更优雅的解决方案。

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 toNULL.

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.
 * 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;
        ArrayList<TreeLinkNode> list1 = new ArrayList<TreeLinkNode>();
        ArrayList<TreeLinkNode> list2 = new ArrayList<TreeLinkNode>();
        
        list1.add(root);
        
        while (!list1.isEmpty()) {
            for (int i = 0; i < list1.size(); i++) {
                if (list1.get(i).left != null) {
                    list2.add(list1.get(i).left);
                }
                if (list1.get(i).right != null) {
                    list2.add(list1.get(i).right);
                }              
            }
        for (int j = 0; j < list2.size() - 1; j++) {
    	    list2.get(j).next = list2.get(j+1);
	    }
            ArrayList<TreeLinkNode> temp = list1;
            list1 = list2;
            list2 = temp; 
            list2.clear();
        }
    }
}

Another solution from Leetcode.

Here is a more elegant solution. The trick is to re-use the populated nextRight pointers. As mentioned earlier, we just need one more step for it to work. Before we passed the leftChild and rightChild to the recursion function itself, we connect the rightChild’s nextRight to the current node’s nextRight’s leftChild. In order for this to work, the current node’s nextRight pointer must be populated, which is true in this case.

void connect(Node* p) {
  if (!p) return;
  if (p->leftChild)
  p->leftChild->nextRight = p->rightChild;
  if (p->rightChild)
    p->rightChild->nextRight = (p->nextRight) ?
                               p->nextRight->leftChild :
                               NULL;
  connect(p->leftChild);
  connect(p->rightChild);
}