Lintcode 二叉树前序遍历

二叉树前序遍历

给出一棵二叉树，返回其节点值的前序遍历。

样例

给出一棵二叉树 {1,#,2,3},

   1
    \
     2
    /
   3

 返回 [1,2,3].

二叉树的前序遍历：根节点->左子树->右子树。下图的前序遍历结果为：abdefgc。

通过递归方式实现前序遍历的代码如下：

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    ArrayList list = new ArrayList();
    public ArrayList preorderTraversal(TreeNode root) {
        // write your code here
        preorder(root);
        return list;
    }
    private void preorder(TreeNode root){
        if(root == null) return;
        list.add(root.val);
        preorder(root.left);
        preorder(root.right);
    }
}

通过非递归方式实现前序遍历的代码如下：

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList preorderTraversal(TreeNode root) {
        // write your code here
        Stack stack = new Stack();
        ArrayList list = new ArrayList();
        while(root!=null||stack.size()>0){
            while(root!=null){
                list.add(root.val);
                stack.push(root);
                root = root.left;
            }
            if(stack.size()>0){
                root = stack.pop();
                root = root.right;
            }  
        }
        return list;
    }
}