404. 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.

Solution:

Method 1: Iterative

We use pre-order traversal.

At any node, if we want to push its left child to the stack, we check if this left child's left child is a left leaf.

If it is, we add its value.

We push a right child into the stack only if this right child has child. (There might be left leaf in this subtree.)



Code:

/**
 * 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) {
        int sum = 0;
        if (root == null) {
            return sum;
        }
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            root = stack.pop();
            if (root.right != null) {
                if (root.right.left != null || root.right.right != null) {
                    stack.push(root.right);
                }
            }
            if (root.left != null) {
                if (root.left.left == null && root.left.right == null) {
                    sum += root.left.val;
                }
                else {
                    stack.push(root.left);
                }
            }
        }
        return sum;
    }
}

Method 2: Recursive

From root, we sum up all its left subtree's left leaves.

Then we add its right subtree's left leaves.



Code:

/**
 * 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 sum = 0;
        if (root.left != null) {
            if (root.left.left == null && root.left.right == null) {
                sum += root.left.val;
            }
            else {
                sum += sumOfLeftLeaves(root.left);
            }
        }
        sum += sumOfLeftLeaves(root.right);
        return sum;
    }
}