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.

M1: recursive

time: O(n), space: O(height)

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

M2: iterative

对于每一个节点，检查它的left child是不是leaf，如果是就加到sum中，如果不是就把left child加入stack。

对于每一个节点的right child，如果它不是leaf就加入stack中

time: O(n), space: O(N)　　-- 最多一层节点的个数

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