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website/content/ChapterFour/0102.Binary-Tree-Level-Order-Traversal.md

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+# [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/)
+
+## 题目
+
+
+Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
+
+**For Example**:
+
+Given binary tree [3,9,20,null,null,15,7],
+
+```
+
+    3
+   / \
+  9  20
+    /  \
+   15   7
+
+```
+
+return its level order traversal as:
+
+```
+
+[
+  [3],
+  [9,20],
+  [15,7]
+]
+
+```
+ 
+
+## 题目大意
+
+按层序从上到下遍历一颗树。
+
+## 解题思路
+
+用一个队列即可实现。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+// 解法一 BFS
+func levelOrder(root *TreeNode) [][]int {
+	if root == nil {
+		return [][]int{}
+	}
+	queue := []*TreeNode{}
+	queue = append(queue, root)
+	curNum, nextLevelNum, res, tmp := 1, 0, [][]int{}, []int{}
+	for len(queue) != 0 {
+		if curNum > 0 {
+			node := queue[0]
+			if node.Left != nil {
+				queue = append(queue, node.Left)
+				nextLevelNum++
+			}
+			if node.Right != nil {
+				queue = append(queue, node.Right)
+				nextLevelNum++
+			}
+			curNum--
+			tmp = append(tmp, node.Val)
+			queue = queue[1:]
+		}
+		if curNum == 0 {
+			res = append(res, tmp)
+			curNum = nextLevelNum
+			nextLevelNum = 0
+			tmp = []int{}
+		}
+	}
+	return res
+}
+
+// 解法二 DFS
+func levelOrder1(root *TreeNode) [][]int {
+	levels := [][]int{}
+	dfsLevel(root, -1, &levels)
+	return levels
+}
+
+func dfsLevel(node *TreeNode, level int, res *[][]int) {
+	if node == nil {
+		return
+	}
+	currLevel := level + 1
+	for len(*res) <= currLevel {
+		*res = append(*res, []int{})
+	}
+	(*res)[currLevel] = append((*res)[currLevel], node.Val)
+	dfsLevel(node.Left, currLevel, res)
+	dfsLevel(node.Right, currLevel, res)
+}
+
+```