20190508

code/lc99.java

+package code;
+
+import java.util.Stack;
+/*
+ * 99. Recover Binary Search Tree
+ * 题意：二叉搜索树中两个节点错位了，恢复二叉搜索树，用O(1)空间
+ * 难度：Hard
+ * 分类：Tree, Depth-first Search
+ * 思路：只要记录错乱的节点就可以了，最后交换两个节点的值
+ *      先序遍历  在print那个地方加上逻辑代码
+ * Tips：
+ */
+public class lc99 {
+    public class TreeNode {
+        int val;
+        TreeNode left;
+        TreeNode right;
+        TreeNode(int x) {
+            val = x;
+        }
+    }
+
+    TreeNode tn1 = null;
+    TreeNode tn2 = null;
+    public void recoverTree(TreeNode root) {
+        inorder(root);
+        int temp = tn1.val;
+        tn1.val = tn2.val;
+        tn2.val = temp;
+    }
+
+    public void inorder(TreeNode root){
+        TreeNode pre = null;
+        Stack<TreeNode> st = new Stack();
+        while(root!=null || !st.isEmpty()){
+            while(root!=null){
+                st.add(root);
+                root = root.left;
+            }
+            root = st.pop();
+            if(pre!=null && pre.val>root.val && tn1==null)
+                tn1 = pre;
+            if(pre!=null && pre.val>root.val && tn1!=null)   //可能被执行多次  eg  3 2 1  tn2 先被赋值2 后被赋值1
+                tn2 = root;
+            pre = root;
+            root = root.right;
+        }
+    }
+}