a116

README.md

@@ -60,6 +60,8 @@ leetcode 题解
 
 [110. Balanced Binary Tree](Solution/101-150/110.md)
 
+[116. Populating Next Right Pointers in Each Node](Solution/101-150/116.md)
+
 [119. Pascal's Triangle II](Solution/101-150/119.md)
 
 [121. Best Time to Buy and Sell Stock](Solution/101-150/121.md)

Solution/101-150/116.md

@@ -0,0 +1,34 @@
+#116. Populating Next Right Pointers in Each Node
+[题目链接](https://leetcode.com/problems/populating-next-right-pointers-in-each-node/?tab=Description)
+```java
+public class Solution {
+    public void connect(TreeLinkNode root) {
+        if (root == null) {
+            return;
+        }
+        // 利用队列的性质，采用广度优先便利
+        LinkedList<TreeLinkNode> ll = new LinkedList<>();
+        ll.add(root);
+        // 以层为单位进行循环
+        while (ll.size() != 0) {
+            int n = ll.size();
+            TreeLinkNode node = ll.pop();
+            // 层内循环
+            for (int i = 0; i < n; i++) {
+                if (node.left != null) {
+                    ll.addLast(node.left);
+                    ll.addLast(node.right);
+                }
+                // 如果是层的最后一个，next设为null
+                if (i < n - 1) {
+                    TreeLinkNode nextNode = ll.pop();
+                    node.next = nextNode;
+                    node = nextNode;
+                } else {
+                    node.next = null;
+                }
+            }
+        }
+    }
+}
+```