Lintcode 二叉樹前序遍歷

二叉樹前序遍歷

給出一棵二叉樹，返回其節點值的前序遍歷。

樣例

給出一棵二叉樹 {1,#,2,3},

   1
    \
     2
    /
   3

 返回 [1,2,3].

二叉樹的前序遍歷：根節點->左子樹->右子樹。下圖的前序遍歷結果爲：abdefgc。

通過遞歸方式實現前序遍歷的代碼如下：

/**
 * 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<Integer> list = new ArrayList<Integer>();
    public ArrayList<Integer> 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);
    }
}

通過非遞歸方式實現前序遍歷的代碼如下：

/**
 * 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<Integer> preorderTraversal(TreeNode root) {
        // write your code here
        Stack<TreeNode> stack = new Stack<TreeNode>();
        ArrayList<Integer> list = new ArrayList<Integer>();
        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;
    }
}