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Tree - 102. Binary Tree Level Order Traversal

Tree - 102. Binary Tree Level Order Traversal

102. Binary Tree Level Order Traversal

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]
]

思路:

层序遍历二叉树,这一题和637题很像,637是求每一层的平均值,这一题只让打印出来,可以用bfs做,也可以使用dfs做,这里用bfs做,有递归写法和非递归写法,非递归写法使用了队列在存储下一层要遍历的节点。

代码:

go:

/**

 * Definition for a binary tree node.

 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
/*// bfs recursive
func levelOrder(root *TreeNode) [][]int {
    var res [][]int
    if root == nil { return res }
    
    traversalNode(root, &res, 0)
    
    return res
}

func traversalNode(node * TreeNode, res *[][]int, h int) {
    if node == nil { return }
    if len(*res) < h + 1 {
        *res = append(*res, []int{})
    }
    
    (*res)[h] = append((*res)[h], node.Val)
    traversalNode(node.Left, res, h+1)
    traversalNode(node.Right, res, h+1)
}*/


// bfs iterative
func levelOrder(root *TreeNode) [][]int {
    var res [][]int
    if root == nil {
        return res
    }
    queue := []*TreeNode{root}
    
    for len(queue) != 0 {
        var temp  []int
        var nextQ []*TreeNode
        
        for len(queue) != 0 {
            cur := queue[0]
            queue = queue[1:]
            temp = append(temp, cur.Val)
            if cur.Left != nil {
                nextQ = append(nextQ,cur.Left)
            }
            
            if cur.Right != nil {
                nextQ = append(nextQ,cur.Right)
            }
        }
        
        res = append(res, temp)
        queue = append(queue, nextQ...)        
    }
    
    return res
}

标题:Tree - 102. Binary Tree Level Order Traversal
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/08/16/1565969865025.html

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