【发布时间】:2020-04-10 19:47:19
【问题描述】:
我正在尝试找到二叉树的最小深度;但是,我在示例 5 中的测试用例失败了。我不确定我的逻辑中的缺陷是否适用于所有测试用例。我正在做的一个例子如下:
Example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its minimum:
depth = 2
我有以下代码来完成这个:
class TreeNode {
constructor(val) {
this.val = val;
this.left = this.right = null;
}
}
const minDepth = root => {
if (!root) return 0
const traverse = root => {
let counter = 1
if (!root) return counter
let current
let queue = [root, 's']
while (queue.length > 1) {
current = queue.shift()
if (current === 's') counter++, queue.push('s')
if (!current.left && !current.right) return counter
else {
if (current.left) queue.push(current.left)
if (current.right) queue.push(current.right)
}
}
return counter
}
return root.left && root.right ? Math.min(traverse(root.left), traverse(root.right)) + 1 : traverse(root)
}
//example 1
const tree1 = new TreeNode(3)
tree1.left = new TreeNode(9)
tree1.right = new TreeNode(20)
tree1.right.left = new TreeNode(15)
tree1.right.right = new TreeNode(7)
//example 2
const tree2 = new TreeNode(1)
tree2.left = new TreeNode(2)
tree2.right = new TreeNode(3)
tree2.left.left = new TreeNode(4)
tree2.right.right = new TreeNode(5)
//example 3
const tree3 = new TreeNode(0)
//example 4
const tree4 = new TreeNode(1)
tree4.left = new TreeNode(2)
//example 5 not working
const tree5 = new TreeNode(1)
tree5.left = new TreeNode(2)
tree5.left.right = new TreeNode(3)
tree5.left.right.right = new TreeNode(4)
tree5.left.right.right.right = new TreeNode(5)
console.log(minDepth(tree1))
console.log(minDepth(tree2))
console.log(minDepth(tree3))
console.log(minDepth(tree4))
console.log(minDepth(tree5))
对我缺少什么有什么想法吗?
【问题讨论】:
-
为什么要使用复杂的
traverse函数,其中包含循环、队列和标记值?一个简单的递归函数就足够了。
标签: javascript algorithm binary-tree tree-traversal