程序员面试金典：高度最小的BST、输出单层结点

1.高度最小的BST

题目描述

对于一个元素各不相同且按升序排列的有序序列，请编写一个算法，创建一棵高度最小的二叉查找树。

给定一个有序序列int[] vals,请返回创建的二叉查找树的高度。

import java.util.*;

public class MinimalBST {
    public int buildMinimalBST(int[] vals) {
        if(vals==null||vals.length==0)	return 0;
        return getHigh(vals,0,vals.length-1);
    }
    
    public int getHigh(int []vals,int start,int end){
        if(start>end) return 0;
        int mid=(start+end)/2;
        int lf=getHigh(vals,start,mid-1);
        int lr=getHigh(vals,mid+1,end);
        return lf>lr?lf+1:lr+1;
    }
}

2.输出单层结点

图的宽度优先遍历（BFS），用队列来实现

import java.util.*;

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;
    public TreeNode(int val) {
        this.val = val;
    }
}*/
public class TreeLevel {
    public ListNode getTreeLevel(TreeNode root, int dep) {
		Queue<TreeNode> queue=new LinkedList<TreeNode>();
        ListNode head=new ListNode(-1);
        ListNode node=head;
        queue.offer(root);
        while(!queue.isEmpty()&&dep!=1){
            int size=queue.size();
            for(int i=0;i<size;i++){
                TreeNode temp=queue.poll();
                if(temp.left!=null){
                    queue.offer(temp.left);
                }
                if(temp.right!=null){
                    queue.offer(temp.right);
                }
            }
            dep--;
        }
        
        while(!queue.isEmpty()){
            node.next=new ListNode(queue.poll().val);
            node=node.next;
        }
        return head.next;
    }
}

下面这个和上面的这个差不多，基本没多大区别

import java.util.*;

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;
    public TreeNode(int val) {
        this.val = val;
    }
}*/
public class TreeLevel {
    public ListNode getTreeLevel(TreeNode root, int dep) {
        Queue<TreeNode> queue=new LinkedList<TreeNode>();
		    	ListNode head=new ListNode(-1);
		    	ListNode node=head;
		    	queue.offer(root);
		    	while(!queue.isEmpty()&&dep!=0){
		    		int length=queue.size();
		    		for (int i = 0; i < length; i++) {
						TreeNode tn=queue.poll();
						if(tn.left!=null)
			    			queue.offer(tn.left);
			    		if(tn.right!=null)
			    			queue.offer(tn.right);
			    		
						if(dep==1){
							node.next=new ListNode(tn.val);
							node=node.next;
						}
					}
		    		dep--;
		    	}
		    	return head.next;
    }
}