404. Sum of Left Leaves

Description

Find the sum of all left leaves in a given binary tree.

Example:

tree

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

Solution

DFS

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

BFS

使用一个queue来辅助，

• 对于left node，如果是leaf则直接将其val加到sum上，否则将left node入队列，待后续处理；
• 对于right node，如果是leaf则弃掉，否则将right node入队列，待后续处理。

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