Leetcode - Sum of Left Leaves

My code:
recursion

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int ret = 0;
        if (root.left != null) {
            if (root.left.left == null && root.left.right == null) {
                ret += root.left.val;
            }
            else {
                ret += sumOfLeftLeaves(root.left);
            }
        }
        if (root.right != null) {
            ret += sumOfLeftLeaves(root.right);
        }
        
        return ret;
    }
}

iteration:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.offer(root);
        int ret = 0;
        while (!q.isEmpty()) {
            TreeNode curr = q.poll();
            if (curr.left != null) {
                if (curr.left.left == null && curr.left.right == null) {
                    ret += curr.left.val;
                }
                else {
                    q.offer(curr.left);
                }
            }
            if (curr.right != null) {
                q.offer(curr.right);
            }
        }
        
        return ret;
    }
}

reference:
https://discuss.leetcode.com/topic/60403/java-iterative-and-recursive-solutions/2

竟然没写出来。哎,早点休息了。

Anyway, Good luck, Richardo! -- 10/11/2016

最后编辑于
©著作权归作者所有,转载或内容合作请联系作者
【社区内容提示】社区部分内容疑似由AI辅助生成,浏览时请结合常识与多方信息审慎甄别。
平台声明:文章内容(如有图片或视频亦包括在内)由作者上传并发布,文章内容仅代表作者本人观点,简书系信息发布平台,仅提供信息存储服务。

相关阅读更多精彩内容

友情链接更多精彩内容