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@Geekhyt
Description
递归 dfs
const postorder = function(root) { if (root === null) return [] const res = [] function dfs(root) { if (root === null) return; for (let i = 0; i < root.children.length; i++){ dfs(root.children[i]) } res.push(root.val) } dfs(root) return res }
- 时间复杂度: O(n)
- 空间复杂度: O(n)