LeetCode | Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively?

 

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

 //先序遍历：root->left->right
public class Solution {
    
    public List<Integer> preorderTraversal(ArrayList<Integer> list,TreeNode root){  //重写方法，多态
        if(root==null) return list;
        list.add(root.val);                                       //在内部不能新建List，所有递归遍历结果都写到同一个list
        if(root.left!=null) preorderTraversal(list,root.left);    //注意此处递归调用时的参数list和外围的list是同一个
        if(root.right!=null) preorderTraversal(list,root.right);
        return list;
    }
    
    public List<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();     //List是继承自Collection的接口，而ArrayList和LinkedList才是类
        if(root==null) return result;
        result.add(root.val);
        
        if(root.left!=null) preorderTraversal(result,root.left);  //注意：此处多态调用保证将遍历结果写到result中，如直接调用
        if(root.right!=null) preorderTraversal(result,root.right);//preorderTraversal(root.left),则在调用的内部会新建一个result
                                                                  //而不是将遍历结果写到外围的result中，导致result只有一个值root.val
        return result;       //此处隐含了一个向上转型
    }
}

 

 //利用栈，迭代写法，思想参考中序遍历的博客，只是while循环中重新压入栈的顺序不同
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();      
        if(root==null) return result;
        
        Stack<TreeNode> nodeStack = new Stack<TreeNode>();
        Stack<Integer> countStack = new Stack<Integer>();
        nodeStack.push(root);
        countStack.push(0);
        
        while(!nodeStack.empty()){
            TreeNode node = nodeStack.pop();
            int count = countStack.pop();
            if(count==1){
                result.add(node.val);
            }else{
                if(node.right!=null){
                    nodeStack.push(node.right);
                    countStack.push(0);
                }
                if(node.left!=null){
                    nodeStack.push(node.left);
                    countStack.push(0);
                }
                nodeStack.push(node);
                countStack.push(1);
            }
        }

        return result;      
    }
}