tree:层序遍历

Change-Id: I68fb307fb916f17a03c0640d26572829770f6a1e

Author: qieguo2016

go/leetcode/199.二叉树的右视图.go

@@ -0,0 +1,69 @@
+/*
+ * @lc app=leetcode.cn id=199 lang=golang
+ *
+ * [199] 二叉树的右视图
+ *
+ * https://leetcode-cn.com/problems/binary-tree-right-side-view/description/
+ *
+ * algorithms
+ * Medium (64.85%)
+ * Likes:    478
+ * Dislikes: 0
+ * Total Accepted:    112.5K
+ * Total Submissions: 173.4K
+ * Testcase Example:  '[1,2,3,null,5,null,4]'
+ *
+ * 给定一棵二叉树，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。
+ *
+ * 示例:
+ *
+ * 输入: [1,2,3,null,5,null,4]
+ * 输出: [1, 3, 4]
+ * 解释:
+ *
+ * ⁠  1            <---
+ * ⁠/   \
+ * 2     3         <---
+ * ⁠\     \
+ * ⁠ 5     4       <---
+ *
+ *
+ */
+
+// @lc code=start
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 层次遍历，每层取最后一个节点即可。
+func rightSideView(root *TreeNode) []int {
+	if root == nil {
+		return []int{}
+	}
+	ret := []int{}
+	queue := []*TreeNode{root}
+	for len(queue) > 0 {
+		level := []*TreeNode{}
+		var last int
+		for _, node := range queue {
+			last = node.Val
+			if node.Left != nil {
+				level = append(level, node.Left)
+			}
+			if node.Right != nil {
+				level = append(level, node.Right)
+			}
+		}
+		ret = append(ret, last)
+		queue = level
+	}
+	return ret
+}
+
+// @lc code=end
+