题目描述

给定二叉树的根节点 root ，返回所有左叶子之和。

代码

class Solution {
    
      
public:
    int getLeftSum(TreeNode* node, int& sum) {
    
      
        if (node == nullptr || (node->left==nullptr &&node->right==nullptr)) return 0;

        if (node->left) getLeftSum(node->left, sum);
        if (node->right) getLeftSum(node->right, sum);
        
        //判断左边的叶子节点
        if (node->left != nullptr && (node->left->left == nullptr && node->left->right == nullptr))
            sum += node->left->val;
        return sum;
        
    }
    int sumOfLeftLeaves(TreeNode* root) {
    
      

        int sum = 0;
        return getLeftSum(root, sum);
        
    }
};

class Solution {
    
      
public:
    int sumOfLeftLeaves(TreeNode* root) {
    
      
        if (root == NULL) return 0;
        if (root->left == NULL && root->right== NULL) return 0;

        int leftValue = sumOfLeftLeaves(root->left);    // 左
        if (root->left && !root->left->left && !root->left->right) {
    
       // 左子树就是一个左叶子的情况
            leftValue = root->left->val;
        }
        int rightValue = sumOfLeftLeaves(root->right);  // 右

        int sum = leftValue + rightValue;               // 中
        return sum;
    }
};

class Solution {
    
      
public:
    int sumOfLeftLeaves(TreeNode* root) {
    
      
        if (root == NULL) return 0;
        int leftValue = 0;
        if (root->left != NULL && root->left->left == NULL && root->left->right == NULL) {
    
      
            leftValue = root->left->val;
        }
        return leftValue + sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
    }
}