Leetcode: Sum of Left Leaves

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

Example:

    3
   / \
  9  20
    /  \
   15   7

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

Recursion: 是不是left子数完全由bottom往上第二层决定，如果是left子树且是叶子节点，那么就是left leaves, parent得告诉child是不是在left子树

/**
 * 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) {
        return helper(root, false);
    }
    
    public int helper(TreeNode cur, boolean isLeft) {
        if (cur == null) return 0;
        if (isLeft && cur.left==null && cur.right==null) return cur.val;
        return helper(cur.left, true) + helper(cur.right, false);
    }
}

 BFS:

public class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if(root == null || root.left == null && root.right == null) return 0;
        
        int res = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while(!queue.isEmpty()) {
            TreeNode curr = queue.poll();

            if(curr.left != null && curr.left.left == null && curr.left.right == null) res += curr.left.val;
            if(curr.left != null) queue.offer(curr.left);
            if(curr.right != null) queue.offer(curr.right);
        }
        return res;
    }
}