题解 | 2025/8/5（二叉树中序遍历）（Java）

import java.util.*;
/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

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

    }

}
*/
public class Solution {
    private static TreeNode head;  // 形成的链表头节点
    private static TreeNode pre;   // 记录中序遍历的前一个节点

    public TreeNode Convert(TreeNode pRootOfTree) {
        if (pRootOfTree == null) {
            return null;
        }
        Convert(pRootOfTree.left);
        if (pre == null) {  // 找到最小值
            head = pre = pRootOfTree;
        } else {  // 使得 pre 和 pRootOfTree 构成双向链表
            pre.right = pRootOfTree;
            pRootOfTree.left = pre;
            pre = pRootOfTree;
        }
        Convert(pRootOfTree.right);
        return head;
    }
}