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].

二叉树的前序遍历,根节点→左子树→右子树

解题思路一:

递归实现,JAVA实现如下:

    public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
if (root == null)
return list;
list.add(root.val);
list.addAll(preorderTraversal(root.left));
list.addAll(preorderTraversal(root.right));
return list;
}

解题思路二:

使用stack实现,JAVA实现如下:

	public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
if (root == null)
return list;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
TreeNode pop = root;
while (!stack.isEmpty()) {
pop = stack.pop();
list.add(pop.val);
if (pop.right != null)
stack.add(pop.right);
if (pop.left != null)
stack.add(pop.left);
}
return list;
}

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