Lintcode 二叉树前序遍历

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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<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<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;    }}

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