p116

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LeetCode p116 Populating Next Right Pointers in Each Node 题解

1.题目：

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

题意：

按如上要求改变一个二叉树的数据结构

2.解题思路:

见代码，每行存储最右。

3.代码

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/**
 * 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 HashMap<Integer, TreeLinkNode> d = new HashMap<Integer, TreeLinkNode>();
    public void connect(TreeLinkNode root) {
     if (root == null)
			return;
		root.next = null;
		connect(root, 1);
	}

	public void connect(TreeLinkNode root, int deep) {

		if (root == null)
			return;
		if (d.containsKey(deep)) {
			// System.out.println(deep);
			root.next = d.get(deep);
		} else {
			root.next = null;
		}
		// System.out.println(deep);
		d.put(deep, root);

		connect(root.right, deep + 1);
		connect(root.left, deep + 1);
	}
}

4.一些总结：