[LeetCode]Populating Next Right Pointers in Each Node

标签：java   leetcode   递归   

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

广搜+递归

递归两次

connectLR（）后的树如图

       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;
		connectLR(root);
		connectRL(root);
	}
	
	private void connectLR(TreeLinkNode root){
		if(root.left==null) return;
		root.left.next = root.right;
		connectLR(root.left);
		connectLR(root.right);
	}
	
	private void connectRL(TreeLinkNode root){
		if(root.left==null) return;
		if(root.next!=null){
			root.right.next = root.next.left;
		}
		connectRL(root.left);
		connectRL(root.right);
	}
}

一次递归完成 运行时间：440ms 

/**
 * 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) {
		connect(root,null);
	}
	
	private void connect(TreeLinkNode root, TreeLinkNode brother){
		if(root == null) return;
		root.next = brother;
		connect(root.left,root.right);
		if(brother!=null){
			connect(root.right,brother.left);
		}else{
			connect(root.right,null);
		}
	}
}

[LeetCode]Populating Next Right Pointers in Each Node

标签：java   leetcode   递归   

原文地址：http://blog.csdn.net/guorudi/article/details/41824593