二叉树后序算法（二分）

给定一个数组[2,4,3,6,8,7,5],遍历后使得
此Node->left小于此节点
此Node->right大于此节点

下列代码为后序遍历，时间复杂度为O(N^2)

public class tree{
     
	public static class Node{
     
		public int val;
		public int left;
		public int right;
		public Node(int v){
     val = v;
		}
		
	public static Node arrayToBST(int[] arr){
     
		//array, begin range, end range
		return flatten(arr, 0, arr.length-1);
	}
	
	public static Node flatten(int[] arr, int L, int R){
     
		//We need to check if L is passed by R
		if(L > R){
     return null}

		Node head = new Node(arr[R]);
		
		//if left equals to right that means only one value here
		if(L==R){
     return head;}
		
		//Check where is the middle value, L-1 indicates the tree only have right side, which is the beginning of left side M=0-1+1
		int M = L-1;
		//keep looking to find the middle position if both side equals
		for(int i = L; i<R; ++i){
     
			if(arr[i]<arr[R]){
     
				M = i;
			}
		}
		flatten(arr, L, M);
		flatten(arr, M+1, R-1);
	}
	return head;
}

可以用二分法来进行优化
下列代码时间复杂度为O(logN)

int left = L;
int right = R-1;
int M = L - 1;
while(left<=right){
     
	int mid = left + ((right-left)>>1);
	if(arr[mid]<arr[R]){
     
		M = mid;
		left = mid + 1;
	}else{
     
		right = mid - 1;
	}
}