add 1028

Author: Blankj

README.md

@@ -0,0 +1,80 @@
+package com.blankj.hard._1028;
+
+import com.blankj.structure.TreeNode;
+
+import java.util.LinkedList;
+
+/**
+ * <pre>
+ *     author: Blankj
+ *     blog  : http://blankj.com
+ *     time  : 2020/06/19
+ *     desc  :
+ * </pre>
+ */
+public class Solution {
+
+    public TreeNode recoverFromPreorder(String S) {
+        char[] chars = S.toCharArray();
+        int len = chars.length;
+        LinkedList<TreeNode> stack = new LinkedList<>();
+        for (int i = 0; i < len; ) {
+            int level = 0, val = 0;
+            while (chars[i] == '-') { // 获取所在层级，Character.isDigit() 会比较慢
+                ++i;
+                ++level;
+            }
+            while (i < len && chars[i] != '-') { // 获取节点的值
+                val = val * 10 + chars[i++] - '0';
+            }
+            TreeNode curNode = new TreeNode(val);
+            while (stack.size() > level) { // 栈顶不是父亲，栈顶出栈
+                stack.removeLast();
+            }
+            if (level > 0) {
+                TreeNode parent = stack.getLast();
+                if (parent.left == null) { // 如果节点只有一个子节点，那么保证该子节点为左子节点。
+                    parent.left = curNode;
+                } else {
+                    parent.right = curNode;
+                }
+            }
+            stack.addLast(curNode);
+        }
+        return stack.get(0);
+    }
+
+//    public TreeNode recoverFromPreorder(String S) {
+//        char[] chars = S.toCharArray();
+//        int len = chars.length;
+//        List<TreeNode> levels = new LinkedList<>();
+//        for (int i = 0; i < len; ) {
+//            int level = 0, val = 0;
+//            while (chars[i] == '-') { // 获取所在层级，Character.isDigit() 会比较慢
+//                ++i;
+//                ++level;
+//            }
+//            while (i < len && chars[i] != '-') { // 获取节点的值
+//                val = val * 10 + chars[i++] - '0';
+//            }
+//            TreeNode curNode = new TreeNode(val);
+//            if (level > 0) {
+//                TreeNode parent = levels.get(level - 1);
+//                if (parent.left == null) { // 如果节点只有一个子节点，那么保证该子节点为左子节点。
+//                    parent.left = curNode;
+//                } else {
+//                    parent.right = curNode;
+//                }
+//            }
+//            levels.add(level, curNode); // 因为是前序遍历（根-左-右），也就是右覆盖左时，此时左树已遍历完成，故无需考虑覆盖问题
+//        }
+//        return levels.get(0);
+//    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        TreeNode.print(solution.recoverFromPreorder("1-2--3--4-5--6--7"));
+        System.out.println("==============================================");
+        TreeNode.print(solution.recoverFromPreorder("1-2--3---4-5--6---7"));
+    }
+}