[LeetCode 117] Populating Next Right Pointers in Each Node II

题目链接：populating-next-right-pointers-in-each-node-ii

相同题型：Populating Next Right Pointers in Each Node

import java.util.LinkedList;

/**
 * 
		Follow up for problem "Populating Next Right Pointers in Each Node".
		
		What if the given tree could be any binary tree? Would your previous solution still work?
		
		Note:
		
		You may only use constant extra space.
		For example,
		Given the following binary tree,
		         1
		       /  		      2    3
		     / \    		    4   5    7
		After calling your function, the tree should look like:
		         1 -> NULL
		       /  		      2 -> 3 -> NULL
		     / \    		    4-> 5 -> 7 -> NULL
 *
 */

public class PopulatingNextRightPointersInEachNodeII {

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

		TreeLinkNode(int x) {
			val = x;
		}
	}
	//解法一 递归法
//	61 / 61 test cases passed.
//	Status: Accepted
//	Runtime: 290 ms
//	Submitted: 0 minutes ago

	//时间复杂度：O(n) 空间复杂度：O(1)
    public void connect(TreeLinkNode root) {
		if (root != null) {
			TreeLinkNode firstNode = null;	 //下一层的首个节点		
			TreeLinkNode nextNode = null;	 //下一层已链接的尾节点
			while (root != null) {
				if (root.left != null) {
					if (firstNode == null) {
						firstNode = root.left;
						nextNode = firstNode;
					} else {
						nextNode.next = root.left;
						nextNode = nextNode.next;
					}
				}
				if (root.right != null) {
					if (firstNode == null) {
						firstNode = root.right;
						nextNode = firstNode;
					} else {
						nextNode.next = root.right;
						nextNode = nextNode.next;
					}
				}
				root = root.next;
			}
			connect(firstNode);
		}
    }
   
    
      //解法二 迭代法
//    61 / 61 test cases passed.
//    Status: Accepted
//    Runtime: 271 ms
//    Submitted: 0 minutes ago
  //时间复杂度：O(n) 空间复杂度：O(1)
    public void connect1(TreeLinkNode root) {
		while (root != null) {
			TreeLinkNode firstNode = null;	 //下一层的首个节点		
			TreeLinkNode nextNode = null;	 //下一层已链接的尾节点
			while (root != null) {
				if (root.left != null) {
					if (firstNode == null) {
						firstNode = root.left;
						nextNode = firstNode;
					} else {
						nextNode.next = root.left;
						nextNode = nextNode.next;
					}
				}
				if (root.right != null) {
					if (firstNode == null) {
						firstNode = root.right;
						nextNode = firstNode;
					} else {
						nextNode.next = root.right;
						nextNode = nextNode.next;
					}
				}
				root = root.next;
			}
			root = firstNode;
		}
    }
    
//    解法三：层次遍历法
//    61 / 61 test cases passed.
//    Status: Accepted
//    Runtime: 359 ms
//    Submitted: 0 minutes ago

    //时间复杂度：O(n) 空间复杂度：O(n)
    public void bfs(TreeLinkNode root) {
    	if(root == null) {
    		return;
    	}
    	LinkedList<TreeLinkNode> stack = new LinkedList<TreeLinkNode>();
    	stack.add(root);
    	
    	while(!stack.isEmpty()) {
    		int size = stack.size();
    		for (int i = 0; i < size; i++) {
				TreeLinkNode node = stack.remove(0);
				node.next = (i == size - 1) ? null : stack.getFirst();
				if(node.left != null)
					stack.add(node);
				if(node.right != null)
					stack.add(node.right);
			}
    	}
    }
    
    
	public static void main(String[] args) {
		// TODO Auto-generated method stub

	}

}

[LeetCode 117] Populating Next Right Pointers in Each Node II

原文：http://blog.csdn.net/ever223/article/details/44645947