LeetCode 404. Sum of Left Leaves

Given the root of a binary tree, return the sum of all left leaves.

A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 24
Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively.

Example 2:

Input: root = [1]
Output: 0

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• -1000 <= Node.val <= 1000

---

 最开始想的方法是把int sum和boolean isLeft作为参数传到helper里后来发现用sum的时候没办法做到全局更新待考证于是就改成了sum作为全局变量这时候就不用把int sum作为函数参数了。以及刚开始还踩了个坑就是刚开始的时候我每个if里都return了这样会导致一遇到符合条件的left leaf就立马全部return了就出错了。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int sum = 0;
    public int sumOfLeftLeaves(TreeNode root) {
        helper(root, false);
        return sum;
    }

    private void helper(TreeNode root, boolean isLeft) {
        if (root == null) {
            return;
        }
        if (root.left == null && root.right == null && isLeft) {
            sum += root.val;
        }
        if (root.left != null) {
            helper(root.left, true);
        }
        if (root.right != null) {
            helper(root.right, false);
        }
    }
}

然后看了别人的才意识到其实可以不用全局的sum直接在return的时候return左子树的结果和右子树的结果之和如果是叶子就return val。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        return helper(root, false);
    }

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

还有一种直接可以不用helper的办法它的思想在于直接在root分左子节点和右子节点两种情况考虑。如果是左子节点那么如果它是个叶子就加这个点的值如果不是那就加以这个左子节点为根的树的值如果是右子节点那就加以这个右子节点为根的树的值。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
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;
    }
}

---

迭代BFS

最最模板的方法就是用两个queue一个存TreeNode一个存boolean isLeft两个同时add和remove真的很套路。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        int sum = 0;
        if (root == null) {
            return sum;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        Queue<Boolean> isLeft = new LinkedList<>();
        queue.add(root);
        isLeft.add(false);
        while (!queue.isEmpty()) {
            TreeNode node = queue.remove();
            boolean left = isLeft.remove();
            if (left && node.left == null && node.right == null) {
                sum += node.val;
            }
            if (node.left != null) {
                queue.add(node.left);
                isLeft.add(true);
            }
            if (node.right != null) {
                queue.add(node.right);
                isLeft.add(false);
            }
        }
        return sum;
    }
}

但是还看到了可以不用isLeft queue的方法就是直接判断左子节点是不是叶子。如果是叶子就+不是叶子就把它add进queue。

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