java中求二叉树的所有节点值的汇总的算法

[shengyh]

public BinaryTreeNode(String as_ClientID,
BigDecimal aBd_BS,
String as_LeftID,
String as_RightID)
结构如上。
1
2 3
4 5 6 7
对应的aBd_BS，ClientID =1, aBd_BS=1
ClientID =2, aBd_BS=2
ClientID =3, aBd_BS=3
ClientID =4, aBd_BS=4
方法：getAllBS(ClientID, 该二叉树的HashMap结构)
HashMap的Key是ClientID，value是BinaryTreeNode。
如何拿到clientID下所有的sumBS？getAllBS方法传的参数也可以变。

[gzhua99]

递归的做

[whyxx]

遍历一把,加起来不就行了.

[bluesmile979]

临时写的，凑合看吧。不好意思了

BinaryTreeNode->BTN


public void getAllBS( HashMap result, String root_clientid ){
BTN root = hashmap.get(root_clientid)
if( root == null )return
String kleft = root.as_LeftID;
if( kleft != null ){
BTn vleft = hashmap.get( kleft );
if( vleft!= null ){
result.put( kleft, vleft );
getAllBS( kleft );
}
}
String kright = root.as_RightID;
if( kright != null ){
BTn vright = hashmap.get( kright );
if( vright!= null ){
result.put( kright, vright );
getAllBS( kright )
}
}

}

[bluesmile979]

public void getAllBS( HashMap result, String root_clientid ){
BTN root = hashmap.get(root_clientid)
if( root == null )return
String kleft = root.as_LeftID;
if( kleft != null ){
BTn vleft = hashmap.get( kleft );
if( vleft!= null ){
result.put( kleft, vleft );
getAllBS( result, kleft );
}
}
String kright = root.as_RightID;
if( kright != null ){
BTn vright = hashmap.get( kright );
if( vright!= null ){
result.put( kright, vright );
getAllBS( result, kright )
}
}

}