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ddd375af20
* Update copyright * Update the Python code * Fix the code comments in ArrayBinaryTree * Fix the code comments in ArrayBinaryTree * Roll back time_comlexity.py * Add the visualizing code(pythontutor) blocks to the chapter complexity, data structure, array and linked list, stack and queue, hash table, and backtracking * Fix the code comments
101 lines
2 KiB
Go
101 lines
2 KiB
Go
// File: array_binary_tree.go
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// Created Time: 2023-07-24
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// Author: Reanon (793584285@qq.com)
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package chapter_tree
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/* 数组表示下的二叉树类 */
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type arrayBinaryTree struct {
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tree []any
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}
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/* 构造方法 */
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func newArrayBinaryTree(arr []any) *arrayBinaryTree {
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return &arrayBinaryTree{
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tree: arr,
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}
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}
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/* 列表容量 */
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func (abt *arrayBinaryTree) size() int {
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return len(abt.tree)
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}
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/* 获取索引为 i 节点的值 */
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func (abt *arrayBinaryTree) val(i int) any {
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// 若索引越界,则返回 null ,代表空位
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if i < 0 || i >= abt.size() {
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return nil
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}
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return abt.tree[i]
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}
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/* 获取索引为 i 节点的左子节点的索引 */
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func (abt *arrayBinaryTree) left(i int) int {
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return 2*i + 1
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}
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/* 获取索引为 i 节点的右子节点的索引 */
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func (abt *arrayBinaryTree) right(i int) int {
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return 2*i + 2
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}
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/* 获取索引为 i 节点的父节点的索引 */
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func (abt *arrayBinaryTree) parent(i int) int {
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return (i - 1) / 2
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}
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/* 层序遍历 */
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func (abt *arrayBinaryTree) levelOrder() []any {
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var res []any
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// 直接遍历数组
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for i := 0; i < abt.size(); i++ {
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if abt.val(i) != nil {
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res = append(res, abt.val(i))
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}
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}
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return res
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}
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/* 深度优先遍历 */
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func (abt *arrayBinaryTree) dfs(i int, order string, res *[]any) {
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// 若为空位,则返回
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if abt.val(i) == nil {
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return
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}
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// 前序遍历
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if order == "pre" {
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*res = append(*res, abt.val(i))
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}
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abt.dfs(abt.left(i), order, res)
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// 中序遍历
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if order == "in" {
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*res = append(*res, abt.val(i))
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}
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abt.dfs(abt.right(i), order, res)
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// 后序遍历
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if order == "post" {
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*res = append(*res, abt.val(i))
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}
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}
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/* 前序遍历 */
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func (abt *arrayBinaryTree) preOrder() []any {
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var res []any
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abt.dfs(0, "pre", &res)
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return res
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}
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/* 中序遍历 */
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func (abt *arrayBinaryTree) inOrder() []any {
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var res []any
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abt.dfs(0, "in", &res)
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return res
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}
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/* 后序遍历 */
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func (abt *arrayBinaryTree) postOrder() []any {
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var res []any
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abt.dfs(0, "post", &res)
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return res
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}
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