LeetCode | Populating Next Right Pointers in Each Node

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

该题有递归（解法1）和非递归（解法2）的写法，明显非递归的写法才能满足题目要求的O(1)的空间复杂度。

解法1

public class PopulatingNextRightPointersInEachNode {
	public class TreeLinkNode {
		int val;
		TreeLinkNode left, right, next;

		TreeLinkNode(int x) {
			val = x;
		}
	}

	public void connect(TreeLinkNode root) {
		if (root == null) {
			return;
		}
		if (root.left != null) {
			root.left.next = root.right;
		}
		if (root.right != null && root.next != null) {
			root.right.next = root.next.left;
		}
		connect(root.left);
		connect(root.right);
	}
}

public class PopulatingNextRightPointersInEachNode {
	public class TreeLinkNode {
		int val;
		TreeLinkNode left, right, next;

		TreeLinkNode(int x) {
			val = x;
		}
	}

	public void connect(TreeLinkNode root) {
		TreeLinkNode leftMostNode = root;
		while (leftMostNode != null) {
			TreeLinkNode node = leftMostNode;
			while (node != null) {
				if (node.left != null) {
					node.left.next = node.right;
				}
				if (node.right != null && node.next != null) {
					node.right.next = node.next.left;
				}
				node = node.next;
			}
			leftMostNode = leftMostNode.left;
		}
	}
}

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LeetCode | Populating Next Right Pointers in Each Node

原文：http://blog.csdn.net/perfect8886/article/details/19932463