【剑指Offer】36. 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 {
    // pre->cur,head为链表头。
    TreeNode head, pre;
    public TreeNode Convert(TreeNode root) {
        dfs(root);
        return head;        
    }
    private void dfs(TreeNode cur) {
        // 终止条件
        if(cur == null) return;
        
        // 左节点
        dfs(cur.left);
        // 根节点
        // 如果pre遍历到末端
        if(pre != null) pre.right = cur;
        else head = cur;
        cur.left = pre;
        pre = cur;
        // 右节点
        dfs(cur.right);
    }
}