Formatted question description: https://leetcode.ca/all/404.html

404. Sum of Left Leaves (Easy)

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.

Related Topics:
Tree

Solution 1. DFS

// OJ: https://leetcode.com/problems/sum-of-left-leaves
// Time: O(N)
// Space: O(H)
class Solution {
private:
    int dfs(TreeNode *root, bool isLeftChild) {
        if (!root) return 0;
        if (!root->left && !root->right && isLeftChild) return root->val;
        return dfs(root->left, true) + dfs(root->right, false);
    }
public:
    int sumOfLeftLeaves(TreeNode* root) {
        return dfs(root, false);
    }
};

Java

/**
 * 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 || root.left == null && root.right == null)
            return 0;
        int sum = 0;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            TreeNode left = node.left, right = node.right;
            if (isLeaf(left))
                sum += left.val;
            if (left != null)
                queue.offer(left);
            if (right != null)
                queue.offer(right);
        }
        return sum;
    }

    public boolean isLeaf(TreeNode node) {
        if (node == null)
            return false;
        if (node.left == null && node.right == null)
            return true;
        return false;
    }
}

// OJ: https://leetcode.com/problems/sum-of-left-leaves
// Time: O(N)
// Space: O(H)
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root, bool isLeft = false) {
        if (!root) return 0;
        if (!root->left && !root->right) return isLeft ? root->val : 0;
        return sumOfLeftLeaves(root->left, true) + sumOfLeftLeaves(root->right);
    }
};

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def sumOfLeftLeaves(self, root):
    """
    :type root: TreeNode
    :rtype: int
    """

    def helper(root, isLeft):
      if not root:
        return None
      left = helper(root.left, True)
      right = helper(root.right, False)
      ret = 0
      if left is None and right is None and isLeft:
        return root.val
      if left:
        ret += left
      if right:
        ret += right
      return ret

    ret = helper(root, False)
    if ret:
      return ret
    return 0

404 - Sum of Left Leaves

404. Sum of Left Leaves (Easy)

Solution 1. DFS

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