多叉树到指定节点的路径

基本上是使用深度优先遍历的套路,以下方法获取的是沿途的所有节点

export function getPathNodesByKey(root,stack,fCompare) {
    let b = false;
    if (root != null) {  
        stack.push(root);  
        if(fCompare(root)){return true}
        var children = root.children;
        if(children){
            for (var i = 0; i < children.length; i++){
                b = getPathByKey(children[i],stack,fCompare);  
                if(b){
                    break;
                }
            } 
        }  
        if(!b){
            stack.pop();
        } 
    }  
    return b;
}

export function getAllPathNodesByKey(root,stack,fCompare) {
    let b = false;
    if (root != null) {  
        stack.push(root);  
        if(fCompare(root)){return true}
        var children = root.children;
        if(children){
            for (var i = 0; i < children.length; i++){
                b = getAllPathByKey(children[i],stack,fCompare);
            } 
        }  
        if(!b){
            stack.pop();
        } 
    }  
    return b;
}

获取沿途所有路径

function getAllPathByKey(root,stack,fCompare,res) {
    if(root == null){
        return;
    }
    stack.push(root);  
    if(fCompare(root)){
        res.push(JSON.parse(JSON.stringify(stack)));
        return;
    }
    var children = root.children;
    if(children){
        for (var i = 0; i < children.length; i++){
            getAllPathByKey(children[i],stack,fCompare,res);  
            stack.pop();
        } 
    }  
}

function getFisrtPathByKey(root,stack,fCompare,res) {
    if(root == null){
        return;
    }
    stack.push(root);  
    if(fCompare(root)){
        res.push(JSON.parse(JSON.stringify(stack)));
        return true;
    }
    var children = root.children;
    if(children){
        for (var i = 0; i < children.length; i++){
            let hasNode = getFisrtPathByKey(children[i],stack,fCompare,res);  
            stack.pop();
            if(hasNode){
                return true;
            }
        } 
    }  
}

posted @ 2020-01-17 10:53  全玉  阅读(1320)  评论(0编辑  收藏  举报