[LeetCode] 108. Convert Sorted Array to Binary Search Tree

Problem

Given a sorted (increasing order) array, Convert it to create a binary tree with minimal height.

Example

Given [1,2,3,4,5,6,7], return

     4
   /   \
  2     6
 / \    / \
1   3  5   7

Note

二分法找到数组的中位数，置为树的root，递归找到前半段和后半段的中位数，分别置为左右子树。直到start = mid或end = mid为止。

Solution

Recursive

public class Solution {
    public TreeNode sortedArrayToBST(int[] A) {  
        return helper(0, A.length - 1, A);
    }  
    public TreeNode helper(int start, int end, int[]A) {
        if (start > end) return null;
        int mid = start + (end - start) / 2;
        TreeNode root = new TreeNode(A[mid]);
        root.left = helper(start, mid - 1, A);
        root.right = helper(mid + 1, end, A);
        return root;
    }
}

Iterative

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums == null || nums.length == 0) return null;
        TreeNode head = new TreeNode(0);
        
        Deque<TreeNode> nodeStack = new LinkedList<>();
        Deque<Integer> leftIndexStack = new LinkedList<>();
        Deque<Integer> rightIndexStack = new LinkedList<>();
        
        nodeStack.push(head);
        leftIndexStack.push(0);
        rightIndexStack.push(nums.length-1);
        
        while (!nodeStack.isEmpty()) {
            TreeNode curNode = nodeStack.pop();
            int left = leftIndexStack.pop();
            int right = rightIndexStack.pop();
            int mid = left+(right-left)/2;
            curNode.val = nums[mid];
            
            if (left < mid) {
                curNode.left = new TreeNode(0);
                nodeStack.push(curNode.left);
                leftIndexStack.push(left);
                rightIndexStack.push(mid-1);
            }
            
            if (mid < right) {
                curNode.right = new TreeNode(0);
                nodeStack.push(curNode.right);
                leftIndexStack.push(mid+1);
                rightIndexStack.push(right);
            }
        }
        
        return head;
    }
}