求深度优先遍历和广度优先遍历的pascal代码

发布网友发布时间：2022-04-22 13:36

共2个回答

懂视网时间：2022-04-22 17:57

本篇文章给大家带来的内容是关于JavaScript实现DOM树的深度优先遍历和广度优先遍历（代码实例），有一定的参考价值，有需要的朋友可以参考一下，希望对你有所帮助。

深度优先遍历

// 非递归,首次传入的node值为DOM树中的根元素点，即html
// 调用：deep(document.documentElement)
function deep (node) {
 var res = []; // 存储访问过的节点
 if (node != null) {
 var nodeList = []; // 存储需要被访问的节点
 nodeList.push(node);
 while (nodeList.length > 0) {
 var currentNode = nodeList.pop(); // 当前正在被访问的节点
 res.push(currentNode);
 var childrens = currentNode.children;
 for (var i = childrens.length - 1; i >= 0; i--) {
 nodeList.push(childrens[i]);
 }
 }
 }
 return res;
}

// 使用递归
var res = []; // 存储已经访问过的节点
function deep (node) {
 if (node != null) { // 该节点存在
 res.push(node);
 // 使用childrens变量存储node.children，提升性能，不使用node.children.length，从而不必在for循环遍历时每次都去获取子元素
 for (var i = 0, childrens = node.children; i < childrens.length; i++) {
 deep(childrens[i]);
 }
 }
 return res;
}

广度优先遍历

// 递归
var res = [];
function wide (node) {
 if (res.indexOf(node) === -1) {
 res.push(node); // 存入根节点
 }
 var childrens = node.children;
 for (var i = 0; i < childrens.length; i++) {
 if (childrens[i] != null) {
 res.push(childrens[i]); // 存入当前节点的所有子元素
 }
 }
 for (var j = 0; j < childrens.length; j++) {
 wide(childrens[j]); // 对每个子元素递归
 }
 return res;
}

// 非递归
function wide (node) {
 var res = [];
 var nodeList = []; // 存储需要被访问的节点
 nodeList.push(node);
 while (nodeList.length > 0) {
 var currentNode = nodeList.shift(0);
 res.push(currentNode);
 for (var i = 0, childrens = currentNode.children; i < childrens.length; i++) {
 nodeList.push(childrens[i]);
 } 
 }
 return res;
}

热心网友时间：2022-04-22 15:05

var v:array[1..100]of boolean;
a:array[1..100,1..100]of integer;
i,r,j,n:integer;
procere dfs(i:integer);
var j:integer;
begin
v[i]:=true;
write(i:5);
for j:=1 to n do
if (not v[j]) and (a[i,j]=1) then dfs(j);
end;

begin

assign(input,'dfs.in'); reset(input);
assign(output,'dfs.out'); rewrite(output);

read(n);
fillchar(v,sizeof(v),false);
for i:=1 to n do
for j:=1 to n do
read(a[i,j]);
for i:=1 to n do
if not V[i] then dfs(i);

close(input);
close(output);
end. var a:array[0..1000,0..1000] of integer;
i,j,k,n,s:integer;
v:array[0..1000] of boolean;
b:array[0..1000] of integer;
front,tail:integer;

procere bfs(x:integer);
var j:integer;
begin

v[i]:=true;
b[tail]:=x;

while front<=tail do
begin
for j:=1 to n do
if (a[b[front],j]=1) and (v[j]=false) then
begin
inc(tail);
v[j]:=true;
b[tail]:=j;
end;
inc(front);
end;
end;

begin
assign(input,'bfs.in'); reset(input);
assign(output,'bfs.out');rewrite(output);

readln(n); front:=1; for i:=1 to n do
for j:=1 to n do
read(a[i,j]);

for i:=1 to n do
if not v[i] then begin tail:=front; bfs(i); end;
for i:=1 to n do writeln(b[i]);

close(input);
close(output);
end.