177 - 排序数组转为二叉树

4.6

清明前做了一半的题目，今天做出来了。

学到几个小的点，计算log2 的时候 应该是 Math.log(n)/ Math.log(double(2)),还是很麻烦的。

这道题的想法就是，先算出最小的树高，然后就可以找到根节点的位置了。

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param A: an integer array
     * @return: a tree node
     */
   	    public TreeNode sortedArrayToBST(int[] A) {  
	       //TreeNode root = new TreeNode();
	       if(A.length == 0){
	    	   return null;
	       }
	       int low = 0;
	       int height = A.length -1;
	       return sortedArrayToBST1(A,low,height);
	    }
	    public TreeNode sortedArrayToBST1(int[] A,int low,int height) {  
	        //TreeNode root = new TreeNode();
	        if(low > height){// write your code here
	        	return null;
	        }
	        //TreeNode root = new TreeNode();
	        if(height == low){ 
	        	TreeNode root = new TreeNode(A[low]);
	        	return root;
	        }
	        //System.out.println("height:" + height +"  low:" + low);
	        int level = (int)(Math.log(height - low +1)/Math.log((double)2));
	        //System.out.println("level:" + level);//高度减一
	        int flag = (int)Math.pow(2, level)-1;
	        //System.out.println("flag:" + flag);
	        TreeNode root = new TreeNode(A[low + flag]);
	       // System.out.println("root.val:" + root.val);
	        root.left = sortedArrayToBST1(A,low,low + flag-1);
	        root.right = sortedArrayToBST1(A,low + flag+1,height);//根节点的位置
	        return root;
	    }
}