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144. 二叉树的前序遍历 #14

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@Geekhyt

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@Geekhyt

原题链接

172e5c5eb83e4be9

前序遍历:先打印当前节点,再打印当前节点的左子树,最后打印当前节点的右子树 (ABCDEFG)

const preorderTraversal = function(root) {
    const result = [];
    function pushRoot(node){
        if (node !== null) {
            result.push(node.val);
            if (node.left !== null){ 
                pushRoot(node.left);
            }
            if (node.right !== null) {
                pushRoot(node.right);
            } 
        }
    }
    pushRoot(root);
    return result;
};
  • 时间复杂度: O(n)
  • 空间复杂度: O(n)

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