题解 | #二叉搜索树与双向链表#

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    public TreeNode head = null;
    public TreeNode end = null;
    public TreeNode Convert(TreeNode pRootOfTree) {
        invert(pRootOfTree);
        return head;
    }

    private void invert(TreeNode node) {
        if (node == null) return;
        invert(node.left);
        if (end == null) {
            head = node;
            end = node;
        } else {
            // 拼接到end的右侧，建立双向关联
            end.right = node;
            node.left = end;
            end = node;
        }
        invert(node.right);
    }
}

解题思想：递归中序遍历