给定权值(5,10,12,15,30,40)，构造相应的哈夫曼树.要求写出构造步骤

按权值大小排列后 5,10,12,15,30,40
只要按照将最小的两个合并, 合并后的值再入列中(最小的两个出列), 至到列中只有一个值.
得到序列5+10=15， （12，15，15，30，40）
[5]`````[10]
\`````/
\```/`
\`/
` `(15)`
从（12，15，15，30，40）找两个最小的12+15=27
[5]`````[10]````````[12]``````[15]
``\`````/`````````````\``````/
` `\```/` ``````````` `\````/`
````\`/`````````````````\``/
````(15)````` ````````(27)`
依次从新的序列中取2个想加和最小的。
（15，27，30，40）-----取15+27=42；
（30，40，42）---------取30+40=70；
（42，70）-------------取42+70=112；
（112）就剩一个终止。

[5]`````[10]`[12]``````[15]
``\`````/``````\``````/
`0`\```/`1`````0\````/`1
````\`/``````````\``/
````(15)````` （27）``[30]`````[40]
``````\`````` /`````````\``````/
`````0`\```` /`1````````0\````/`1
````````\`` /`````````````\``/
```````` （42)````````````` (70)``````
```````````\`````` ````````````/
``````````0`\``````` ```````/
`````````````\ ```````````/
`````````````(46)``````````1/`
````````````````\``````````/``
```````````````0`\````````/``
``````````````````\``````/``
````````````````````(112)

图真不好画，详细可以参考