My answer: find the root of left and right subtree, and then link them to root. Basically, it is converting each sub-array to a BST
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode sortedArrayToBST(int[] nums) { if (nums == null || nums.length == 0) { return null; } TreeNode root = new TreeNode(nums[(0 + nums.length - 1) / 2]); helper(root, nums, 0, (0 + nums.length - 1) / 2 - 1, (0 + nums.length - 1) / 2 + 1, nums.length - 1); return root; } void helper(TreeNode root, int[] nums, int leftMin, int leftMax, int rightMin, int rightMax) { TreeNode leftRoot = null; TreeNode rightRoot = null; if (leftMin <= leftMax) { leftRoot = new TreeNode(nums[(leftMin + leftMax) / 2]); } if (rightMin <= rightMax) { rightRoot = new TreeNode(nums[(rightMin + rightMax) / 2]); } root.left = leftRoot; root.right = rightRoot; if (leftRoot != null) { helper(leftRoot, nums, leftMin, (leftMin + leftMax) / 2 - 1, (leftMin + leftMax) / 2 + 1, leftMax); } if (rightRoot != null) { helper(rightRoot, nums, rightMin, (rightMin + rightMax) / 2 - 1, (rightMin + rightMax) / 2 + 1, rightMax); } } }
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