利用graphviz 绘制树状图

digraph action {
    node [shape = record,height=.1];

    node0 [label = "<head> head|<body> body|<foot> foot", height=.5]
    node2 [shape = box label="mind"]

    node0:head:n -> node2:n [label = "n"] //node0的头结点的n方位指向node2的n方位
    node0:head:ne -> node2:ne [label = "ne"]
    node0:head:e -> node2:e [label = "e"]
    node0:head:se -> node2:se [label = "se"]
    node0:head:s -> node2:s [label = "s"]
    node0:head:sw -> node2:sw [label = "sw"]
    node0:head:w -> node2:w [label = "w"]
    node0:head:nw -> node2:nw [label = "nw"]
    node0:head:c -> node2:c [label = "c"] //center
    node0:head:_ -> node2:_ [label = "_"] //_表示任意位置

    node0:body[style=filled color=lightblue]
}

digraph {
    rankdir = LR
    splines = false

    A -> B -> C -> D -> F [color = green]
    E -> F -> B -> D [color = blue]
    B -> E -> H[color = red]
}

View Code

digraph G {
    compound = true  // 允许子图间存在边
    ranksep = 1
    node [shape = record]

    subgraph cluster_hardware {
        label = "hardware"
        color = lightblue
        CPU Memory //两小图CPU,Memory从右向左添加
    }

    subgraph cluster_kernel {
        label = "kernel"
        color = green
        Init IPC
    }

    subgraph cluster_libc {
        label = "libc"
        color = yellow
        glibc
    }

    CPU -> Init [lhead = cluster_kernel ltail = cluster_hardware]
    IPC -> glibc [lhead = cluster_libc ltail = cluster_kernel]
}

输出.dot文件然后通过终端输入“dot x.dot -T png -o x.png”输出图形。

所以可以利用Tree root先根遍历，从而输出为.dot文本。

void  process_null_node(int index, int nullcount, FILE* stream)
{
        fprintf(stream, "    null%d [shape=hexagon, width = .2,height = .2, style = filled, color = \"#636e72\", fontcolor = white]\n", nullcount);
        fprintf(stream, "  node%d -> null%d\n",index, nullcount);
}
void __tree2dot(Huffman root,FILE* stream)
{
    static int  null_node_cnt = 0;
    if(root) {
        if(root->ch!=0)
        fprintf( stream, "node%d [ label = \"<f0>^|{<head>\'%c\'|freq =  %d|CODE = %s}|<f2>^\"]\n", root->index, root->ch, root->freq, root->code);
        else fprintf( stream, "node%d [label = \"<f0>|<f1>%d|<f2>\"]\n", root->index, root->freq);
    }
    if(root->left)
    {
        fprintf(stream,"  node%d:f0 -> node%d:n [ label = \"0\"]\n",root->index,root->left->index);
        __tree2dot(root->left,stream);
    }
    else
    {
        //process_null_node(root->index,null_node_cnt++,stream);
    }
    if(root->right)
    {
        fprintf(stream,"  node%d:f2 -> node%d:n [ label = \"1\"]\n",root->index,root->right->index);
        __tree2dot(root->right,stream);
    }
    else
    {
        //process_null_node(root->index,null_node_cnt++,stream);
    }

}
void tree2index(Huffman root){
    if(root==NULL) return;
    queue<Node *> s;
    s.push(root);
    int i = 0;
    while(!s.empty()){
        s.front()->index = i++;
        Node *temp = s.front();
        s.pop();
        if(temp->left)
        s.push(temp->left);
        if(temp->right)
        s.push(temp->right);
    }
}
int tree2dot(Huffman tree,char* filename)
{
    assert(tree != NULL && filename != NULL);
    Tree_array T_array[100]={};
    FILE* stream = fopen(filename,"w+");
    if(stream == NULL) {
        fprintf(stderr, "open failed \n");
        return -1;
    }
    tree2index(tree);
    fprintf(stream,"digraph {\n");
    fprintf(stream,"node [shape=\"record\", height=.1]\n");
    __tree2dot(tree,stream);
    fprintf(stream,"}\n");
    fclose(stream);
    return 0;
}