二叉樹算法之構造

樹（Tree）在數據結構仍是很重要的，這裏表示二叉樹用括號表示法表示。
先寫一個二叉樹節點類：

// 二叉樹節點
class BTNode {
    public $data;

    public $lchild = NULL;

    public $rchild = NULL;

    public function __construct($data) {
        $this->data = $data;
    }
}

而後構造二叉樹：

function CreateBTNode(BTNode &$root = NULL, string $str)
{
    $strArr = str_split($str);
    $stack = [];
    $p = NULL; // 指針
    $top = -1;
    $k = $j = 0;
    $root = NULL;
    foreach ($strArr as $ch) {
        switch ($ch) {
            case '(':
                $top++;
                array_push($stack, $p);
                $k = 1;
                break;
            case ')':
                array_pop($stack);
                break;
            case ',':
                $k = 2;
                break;
            default:
                $p = new BTNode($ch);
                if($root == NULL) {
                    $root = $p;
                } else {
                    switch ($k) {
                        case 1:
                            end($stack)->lchild = $p;
                            break;
                        case 2:
                            end($stack)->rchild = $p;
                            break;
                    }
                }
                break;
        }
    }
}

這裏寫上一個打印二叉樹的函數（中序遍歷）：

function PrintBTNode($node)
{
    if($node != NULL) {
        PrintBTNode($node->lchild);
        echo $node->data;
        PrintBTNode($node->rchild);
    }
}

運行結果：
輸入一個字符串
"A(B(C,D),G(F))"

go語言實現

package main

import (
    "fmt"
    "strings"
)

type BinaryTreeNode struct {
    data   string
    lChild *BinaryTreeNode
    rChild *BinaryTreeNode
}

func CreateBinaryTree(sequence string) *BinaryTreeNode {
    words := strings.Split(sequence, "")
    stack := []*BinaryTreeNode{}
    var p *BinaryTreeNode = nil
    top := -1
    k := 0
    var root *BinaryTreeNode = nil
    for _, word := range words {
        switch word {
        case "(":
            top++
            stack = append(stack, p)
            k = 1
        case ")":
            stack = stack[0:top]
            top--
            k = 0
        case ",":
            k = 2
        default:
            p = &BinaryTreeNode{
                word,
                nil,
                nil,
            }
            if root == nil {
                root = p
            } else {
                endItem := stack[top]
                switch k {
                case 1:
                    endItem.lChild = p
                case 2:
                    endItem.rChild = p
                }
            }
        }
    }
    return root
}

// 中序遍歷
func inOrderBinaryTree(root *BinaryTreeNode) {
    if root != nil {
        inOrderBinaryTree(root.lChild)
        fmt.Print(root.data)
        inOrderBinaryTree(root.rChild)
    }
}

func main() {
    testStr := "A(B(C,D),G(F))"
    root := CreateBinaryTree(testStr)
    inOrderBinaryTree(root)
}