题目链接

    1.       // 中序mirror
    2.      public List<Integer> inorderTraversal(TreeNode root) {
    3.          mirrisPre(root);
    4.          return ansmorris;
    5.      }
    7.      List<Integer> ansmorris = new ArrayList<>();
    8.      public void mirrisPre(TreeNode cur) {
    9.          if(cur == null) return;
    11.          TreeNode mostRight = null;
    12.          while(cur != null) {
    13.              mostRight = cur.left; 
    14.              if(mostRight != null) {
    15.                  while(mostRight.right != null && mostRight.right != cur) { // 如果右边不为空，并且右边不指向后继节点的话就找到其前继节点
    16.                     mostRight = mostRight.right;
    17.                  }
    19.                  if(mostRight.right == null) { // 建立线索指针
    20.                      mostRight.right = cur;
    21.                      // 打印
    22.                     //  ansmorris.add(cur.val);
    23.                      cur = cur.left;
    24.                      continue;
    25.                  } else { // 删除线索指针
    26.                      mostRight.right = null; 
    27.                  }
    29.              } else {
    31.              }
    33.              // 打印节点
    34.              ansmorris.add(cur.val);
    36.              cur = cur.right; // 向右走
    37.          }
    38.      }

    前序遍历

        // 前序mirror
     public List<Integer> inorderTraversal(TreeNode root) {
         mirrisPre(root);
         return ansmorris;
     }

List<Integer> ansmorris = new ArrayList<>();
     public void mirrisPre(TreeNode cur) {
         if(cur == null) return;

         TreeNode mostRight = null;
         while(cur != null) {
             mostRight = cur.left; 
             if(mostRight != null) {
                 while(mostRight.right != null && mostRight.right != cur) { // 如果右边不为空，并且右边不指向后继节点的话就找到其前继节点
                    mostRight = mostRight.right;
                 }

                 if(mostRight.right == null) { // 建立线索指针。。。根据后继节点进行找到开始节点。
                     mostRight.right = cur;
                     // 打印
                     ansmorris.add(cur.val);
                     cur = cur.left;
                     continue;
                 } else { // 删除线索指针
                     mostRight.right = null; 
                 }

             } else {
                  ansmorris.add(cur.val);
             }
             cur = cur.right; // 向右走
         }
     }