Leetcode_144_Binary Tree Preorder Traversal

时间:2022-11-26 18:58:53

本文是在学习中的总结,欢迎转载但请注明出处:http://blog.csdn.net/pistolove/article/details/42876699

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?

思路:

(1)题意为前序遍历二叉树。遍历顺序为根—>左—>右。

(2)考虑到用递归比较简单,本文使用递归的思想进行解决,由于比较简单这里不累赘,详见下方代码。

(3)希望本文对你有所帮助。

算法代码实现如下:

/**
 * @author liqq
 */
public List<Integer> preorderTraversal(TreeNode root) {
	List<Integer> result = new LinkedList<Integer>();

	if (root != null) {
		result.add(root.val);
		pre_order(result, root.left);
		pre_order(result, root.right);
	}

	return result;
}

private void pre_order(List<Integer> result, TreeNode curr) {
	if (curr != null) {
		result.add(curr.val);
		pre_order(result, curr.left);
		pre_order(result, curr.right);
	}
}