LeetCode-404-Left Leaf Sum

The sum of the left leaves

Title description: Calculate the sum of all left leaves of a given binary tree.

Please refer to LeetCode official website for example description.

Source: LeetCode
Link: https://leetcode-cn.com/problems/sum-of-left-leaves/
The copyright belongs to Lingkou Network. For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.

Solution 1: Recursion

First, if the root node root is null or there is only one node, it means that there is no leaf node, and 0 is directly returned;

Otherwise, add a recursive method recursive , there are 2 parameters, which are the left and right child nodes of the current node, flag is the identification of the left and right child nodes, the recursive process is as follows:

• Call the recursive method recursive , the parameters are root left and right child nodes, flag is the corresponding identifier;
• Determine if root in the recursive method null , then return;
• If root has no left and right child nodes and flag identified as the left child node, the value of root ;
• Otherwise, the recursive call recursive This , parameters are root left and right child nodes, Flag corresponding logo.

Finally, the return result is the sum of all left leaf nodes.

import com.kaesar.leetcode.TreeNode;

/**
 * @Author: ck
 * @Date: 2021/9/29 7:33 下午
 */
public class LeetCode_404 {
    /**
     * 叶子之和
     */
    public static int result = 0;

    public static int sumOfLeftLeaves(TreeNode root) {
        if (root == null || (root.left == null && root.right == null)) {
            return 0;
        }
        recursive(root.left, true);
        recursive(root.right, false);
        return result;
    }

    /**
     * 递归方法
     *
     * @param root
     * @param flag true表示是左子节点；false表示是右子节点
     */
    public static void recursive(TreeNode root, boolean flag) {
        if (root == null) {
            return;
        }
        if (root.left == null && root.right == null && flag) {
            result += root.val;
        }
        recursive(root.left, true);
        recursive(root.right, false);
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(3);
        root.left = new TreeNode(9);
        root.right = new TreeNode(20);
        root.right.left = new TreeNode(15);
        root.right.right = new TreeNode(7);

        // 期望返回值: 24
        System.out.println(sumOfLeftLeaves(root));
    }
}

[Daily Message] Lazy people wait for opportunities, hardworking people create opportunities.