这个答案基于@Ravindra 的答案，但具有真正的递归。

function searchTree(element, matchingTitle){
     if(element.title == matchingTitle){
          return element;
     }else if (element.children != null){
          var i;
          var result = null;
          for(i=0; result == null && i < element.children.length; i++){
               result = searchTree(element.children[i], matchingTitle);
          }
          return result;
     }
     return null;
}

然后你可以调用它：

var element = data[0];
var result = searchTree(element, 'randomNode_1');

这是一个迭代解决方案：

var stack = [], node, ii;
stack.push(root);

while (stack.length > 0) {
    node = stack.pop();
    if (node.title == 'randomNode_1') {
        // Found it!
        return node;
    } else if (node.children && node.children.length) {
        for (ii = 0; ii < node.children.length; ii += 1) {
            stack.push(node.children[ii]);
        }
    }
}

// Didn't find it. Return null.
return null;

这是一个使用 Stack 方法的迭代函数，其灵感来自FishBasketGordo 的答案，但利用一些ES2015语法来缩短事情。

由于这个问题已经被查看了很多次，我决定更新我的答案，以提供一个带有参数的函数，使其更加灵活：

function search (tree, value, key = 'id', reverse = false) {
  const stack = [ tree[0] ]
  while (stack.length) {
    const node = stack[reverse ? 'pop' : 'shift']()
    if (node[key] === value) return node
    node.children && stack.push(...node.children)
  }
  return null
}

这样，现在可以传递数据tree本身、需要value搜索的数据以及key可以具有所需值的属性：

search(data, 'randomNode_2', 'title')

最后，我使用的原始答案Array.pop导致在多个匹配的情况下匹配最后一个项目。事实上，有些事情可能真的很令人困惑。受到Superole 评论的启发，我Array.shift现在已经使用了它，所以先进先出行为是默认的。

如果你真的想要旧的后进先出行为，我提供了一个额外的 arg reverse：

search(data, 'randomNode_2', 'title', true)

我的回答灵感来自 FishBasketGordo 的迭代回答。它有点复杂，但也更灵活，您可以拥有多个根节点。

/**searchs through all arrays of the tree if the for a value from a property
 * @param aTree : the tree array
 * @param fCompair : This function will receive each node. It's upon you to define which 
                     condition is necessary for the match. It must return true if the condition is matched. Example:
                        function(oNode){ if(oNode["Name"] === "AA") return true; }
 * @param bGreedy? : us true to do not stop after the first match, default is false
 * @return an array with references to the nodes for which fCompair was true; In case no node was found an empty array
 *         will be returned
*/
var _searchTree = function(aTree, fCompair, bGreedy){
    var aInnerTree = []; // will contain the inner children
    var oNode; // always the current node
    var aReturnNodes = []; // the nodes array which will returned

    // 1. loop through all root nodes so we don't touch the tree structure
    for(keysTree in aTree) {
        aInnerTree.push(aTree[keysTree]);
    }
    while(aInnerTree.length > 0) {
        oNode = aInnerTree.pop();
        // check current node
        if( fCompair(oNode) ){
            aReturnNodes.push(oNode);
            if(!bGreedy){
                return aReturnNodes;
            }
        } else { // if (node.children && node.children.length) {
            // find other objects, 1. check all properties of the node if they are arrays
            for(keysNode in oNode){
                // true if the property is an array
                if(oNode[keysNode] instanceof Array){
                    // 2. push all array object to aInnerTree to search in those later
                    for (var i = 0; i < oNode[keysNode].length; i++) {
                        aInnerTree.push(oNode[keysNode][i]);
                    }
                }
            }
        }
    }
    return aReturnNodes; // someone was greedy
}

最后你可以像这样使用这个函数：

var foundNodes = _searchTree(data, function(oNode){ if(oNode["title"] === "randomNode_3") return true; }, false);
console.log("Node with title found: ");
console.log(foundNodes[0]);

如果你想找到所有具有这个标题的节点，你可以简单地切换 bGreedy 参数：

var foundNodes = _searchTree(data, function(oNode){ if(oNode["title"] === "randomNode_3") return true; }, true);
console.log("NodeS with title found: ");
console.log(foundNodes);

你必须使用递归。

var currChild = data[0];
function searchTree(currChild, searchString){
     if(currChild.title == searchString){
          return currChild;
     }else if (currChild.children != null){
          for(i=0; i < currChild.children.length; i ++){
               if (currChild.children[i].title ==searchString){
                    return currChild.children[i];
               }else{
                    searchTree(currChild.children[i], searchString);
               }
          }
          return null;
     }
     return null;
}