二叉搜索树与双向链表最直观java解法

思路很简单，Convert函数得到的就是转换之后的双向链表头，转化出来的左子树连接到root，再将root连接到右子树，就可以了，运行时间11ms

public class Solution {
    public TreeNode Convert(TreeNode pRootOfTree) {
        if(pRootOfTree == null) return null;

        TreeNode leftHead = Convert(pRootOfTree.left);

        if (leftHead != null) {
            TreeNode leftTail = leftHead;
            while(leftTail.right != null) {
                leftTail = leftTail.right;
            }
            leftTail.right = pRootOfTree;   
            pRootOfTree.left = leftTail;
        }

        TreeNode rightHead = Convert(pRootOfTree.right);
        if (rightHead != null) {
            rightHead.left = pRootOfTree;
            pRootOfTree.right = rightHead;
        }
        return leftHead == null?pRootOfTree:leftHead;
    }
}