有序链表转BST(平衡查找二叉树)


    private static class ListNode {
        public int value;
        public ListNode next = null;
    }

    private static class TreeNode {
        public int value;
        public TreeNode left = null;
        public TreeNode right = null;
    }

    private static TreeNode convertList2Tree(ListNode head, int length) {
        if (head == null || length == 0) return null;
        ListNode leftHead = null, rootHead = null, rightHead = null;
        TreeNode leftRoot = null, root = null, rightRoot = null;
        if (length % 2 == 1) {
            // 1 2 3 [4] 5 6 7
            leftHead = head;
            rootHead = moveList(leftHead, length / 2 + 1);
            rightHead = rootHead.next;
            leftRoot = convertList2Tree(leftHead, length / 2);
            rightRoot = convertList2Tree(rightHead, length / 2);
        } else {
            // 0 1 2 [3] 4 5
            leftHead = head;
            rootHead = moveList(leftHead, length / 2 + 1);
            rightHead = rootHead.next;
            leftRoot = convertList2Tree(leftHead, length / 2);
            rightRoot = convertList2Tree(rightHead, length / 2 - 1);
        }
        root = new TreeNode();
        root.left = leftRoot;
        root.right = rightRoot;
        root.value = rootHead.value;
        return root;
    }
    原文作者:二叉查找树
    原文地址: https://blog.csdn.net/qq_17612199/article/details/52671411
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