Merge pull request #110 from Joshua-Lu/patch-27

更新 0101.对称二叉树 Java版本，递归与迭代
youngyangyang04 · May 14, 2021 · 9fc2f69 · 9fc2f69
2 parents af95c8c + cbf41eb
commit 9fc2f69
Showing 1 changed file with 77 additions and 62 deletions.
diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md
@@ -254,93 +254,108 @@ public:
 ## 其他语言版本
 
 Java：
-```java
-public class N0101 {
+
+```Java
     /**
-     * 解法1：DFS，递归。
+     * 递归法
      */
-    public boolean isSymmetric2(TreeNode root) {
-        if (root == null) {
-            return false;
-        }
-
+    public boolean isSymmetric1(TreeNode root) {
         return compare(root.left, root.right);
     }
 
     private boolean compare(TreeNode left, TreeNode right) {
-        if (left == null && right == null) {
-            return true;
+
+        if (left == null && right != null) {
+            return false;
         }
         if (left != null && right == null) {
             return false;
         }
-        if (left == null && right != null) {
+
+        if (left == null && right == null) {
+            return true;
+        }
+        if (left.val != right.val) {
             return false;
         }
+        // 比较外侧
+        boolean compareOutside = compare(left.left, right.right);
+        // 比较内侧
+        boolean compareInside = compare(left.right, right.left);
+        return compareOutside && compareInside;
+    }
 
-        if (left.val == right.val) {
-            return compare(left.left, right.right) && compare(left.right, right.left);
+    /**
+     * 迭代法
+     * 使用双端队列，相当于两个栈
+     */
+    public boolean isSymmetric2(TreeNode root) {
+        Deque<TreeNode> deque = new LinkedList<>();
+        deque.offerFirst(root.left);
+        deque.offerLast(root.right);
+        while (!deque.isEmpty()) {
+            TreeNode leftNode = deque.pollFirst();
+            TreeNode rightNode = deque.pollLast();
+            if (leftNode == null && rightNode == null) {
+                continue;
+            }
+//            if (leftNode == null && rightNode != null) {
+//                return false;
+//            }
+//            if (leftNode != null && rightNode == null) {
+//                return false;
+//            }
+//            if (leftNode.val != rightNode.val) {
+//                return false;
+//            }
+            // 以上三个判断条件合并
+            if (leftNode == null || rightNode == null || leftNode.val != rightNode.val) {
+                return false;
+            }
+            deque.offerFirst(leftNode.left);
+            deque.offerFirst(leftNode.right);
+            deque.offerLast(rightNode.right);
+            deque.offerLast(rightNode.left);
         }
-
-        return false;
+        return true;
     }
 
     /**
-     * 解法2：DFS，迭代
+     * 迭代法
+     * 使用普通队列
      */
     public boolean isSymmetric3(TreeNode root) {
-        if (root == null) {
-            return false;
-        }
-
-        if (!equal(root.left, root.right)) {
-            return false;
-        }
-
-        Deque<TreeNode> st = new LinkedList<>();
-
-        st.push(root.right);
-        st.push(root.left);
-
-        TreeNode curR = root.right;
-        TreeNode curL = root.left;
-
-        while (!st.isEmpty()) {
-            curL = st.pop();
-            curR = st.pop();
-
-            // 前序，处理
-            if (!equal(curL, curR)) {
-                return false;
+        Queue<TreeNode> deque = new LinkedList<>();
+        deque.offer(root.left);
+        deque.offer(root.right);
+        while (!deque.isEmpty()) {
+            TreeNode leftNode = deque.poll();
+            TreeNode rightNode = deque.poll();
+            if (leftNode == null && rightNode == null) {
+                continue;
             }
-
-            if (curR != null && curL != null) {
-                st.push(curL.right);
-                st.push(curR.left);
-                st.push(curR.right);
-                st.push(curL.left);
+//            if (leftNode == null && rightNode != null) {
+//                return false;
+//            }
+//            if (leftNode != null && rightNode == null) {
+//                return false;
+//            }
+//            if (leftNode.val != rightNode.val) {
+//                return false;
+//            }
+            // 以上三个判断条件合并
+            if (leftNode == null || rightNode == null || leftNode.val != rightNode.val) {
+                return false;
             }
+            // 这里顺序与使用Deque不同
+            deque.offer(leftNode.left);
+            deque.offer(rightNode.right);
+            deque.offer(leftNode.right);
+            deque.offer(rightNode.left);
         }
-
         return true;
     }
 
-    private boolean equal(TreeNode l, TreeNode r) {
-        if (l == null && r == null) {
-            return true;
-        }
-        if (l != null && r == null) {
-            return false;
-        }
-        if (l == null && r != null) {
-            return false;
-        }
-        if (l.val == r.val) {
-            return true;
-        }
-        return false;
-    }
-}
 ```
 
 Python：