ExtJS4.x 如何實現下拉樹菜單

function createSimpleTreePanel(){

//因為每個樹中由N個節點組成的,所以要創建樹,必須創建樹所包含的節點(樹必須有壹個或多個根節點).

var vRoot = new Ext.tree.TreeNode({text: '根節點1'}) ;//創建節點

var vNode1 = new Ext.tree.TreeNode({text: '子節點1'}) ;//創建節點

vNode1.appendChild(new Ext.tree.TreeNode({text: '子節點11'}));

vNode1.appendChild(new Ext.tree.TreeNode({text: '子節點12'}));

var vNode2 = new Ext.tree.TreeNode({text: '子節點1'}) ;//創建節點

vRoot.appendChild(vNode1) ;//在vRoot根節點上創建子節點.

vRoot.appendChild(vNode2) ;//在vRoot根節點上創建子節點.

var vTree = new Ext.tree.TreePanel({//創建壹個用於裝載樹的面板

renderTo: document.body,//綁定在body元素上.

width: 300,

height: 200,

root: vRoot

});

}