面试题-算法：二叉树的前序遍历

今天说下二叉树的前序遍历，先来一颗二叉树熟悉熟悉：

前序遍历：先输出该节点，然后输出左孩子，然后输出右孩子。

public class Tree {

    //定义一颗树

    public static class TreeNode{

        int val;

        //左孩子

        private TreeNode left;

        //右孩子

        private TreeNode right;

        TreeNode(int x){

            val = x;

        }    

}

public static ListpreOrder(TreeNode root){

        //结果

        Listlist =new LinkedList<>();

        //栈

        Stackstack =new Stack<>();

        if(root ==null){

            return list;

        }

        stack.push(root);

        while(!stack.isEmpty()){

            //栈顶元素出栈

            TreeNode treeNode =stack.pop();

            list.add(treeNode.val);

            //左孩子是否存在

            if(treeNode.right!=null){

                stack.push(treeNode.right);

            }

            //右孩子是否存在

            if(treeNode.left!=null){

                stack.push(treeNode.left);

    }

}

return list;

}

public static void main(String[] args) {

        //初始化二叉树

        TreeNode root =new TreeNode(1);

        root.left =new TreeNode(2);

        root.right =new TreeNode(3);

        root.left.left =new TreeNode(4);

        root.left.right =new TreeNode(5);

        //例如:上述初始化的二叉树，前序遍历输出1,2,4,5,3

        Listintegers =preOrder(root);

            System.out.println(integers);

    }

}

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