Merge pull request gzc426#337 from V1ncentzzZ/master

gzc426 · web-flow · commit 79eee5245121 · 2018-12-06T21:49:33.000+08:00
2018/12/04打卡
diff --git a/2018.12.04-leetcode101/。V1ncentzzZ.md b/2018.12.04-leetcode101/。V1ncentzzZ.md
@@ -0,0 +1,43 @@
+101. 对称二叉树
+
+给定一个二叉树，检查它是否是镜像对称的。
+
+例如，二叉树 [1,2,2,3,4,4,3] 是对称的。
+
+    1
+   / \
+  2   2
+ / \ / \
+3  4 4  3
+但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:
+
+    1
+   / \
+  2   2
+   \   \
+   3    3
+   
+```
+  /**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public boolean isSymmetric(TreeNode root) {
+        if(root == null) return true;
+        return symmetricCheck(root.left,root.right);
+    }
+    
+    public boolean symmetricCheck(TreeNode node1,TreeNode node2){
+        if(node1 == null && node2 == null) return true;
+        if(node1 == null || node2 == null) return false;
+        return (node1.val != node2.val) ? false : symmetricCheck(node1.left,node2.right) &&  symmetricCheck(node1.right,node2.left);
+   
+    }
+}
+```
diff --git a/2018.12.05-leetcode102/。V1ncentzzZ.md b/2018.12.05-leetcode102/。V1ncentzzZ.md
@@ -0,0 +1,50 @@
+[102. 二叉树的层次遍历] (https://leetcode-cn.com/problems/binary-tree-level-order-traversal/)
+
+给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。
+
+例如:
+给定二叉树: [3,9,20,null,null,15,7],
+
+    3
+   / \
+  9  20
+    /  \
+   15   7
+返回其层次遍历结果：
+
+[
+  [3],
+  [9,20],
+  [15,7]
+]
+
+
+```
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+ // TODO 解题关键：传递树高度
+class Solution {
+    public List<List<Integer>> levelOrder(TreeNode root) {
+        List<List<Integer>> result = new ArrayList<>();
+        if(root == null) return result;
+        executor(root,result,0);
+        return result;    
+    }
+    
+    public void executor(TreeNode node,List<List<Integer>> result,int high){
+        if(node == null) return;
+        if(result.size() < high+1) result.add(new ArrayList<>());
+        result.get(high).add(node.val);
+        high++;
+        executor(node.left,result,high);
+        executor(node.right,result,high);
+    }
+}
+```
