0102. Binary Tree Level Order Traversal

102. 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 #


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)
}


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