LeetCode404.左叶子之和

题目

法一、广度优先搜索

class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if(root == NULL) return 0;
        if(root->left == NULL && root->right == NULL) return 0;
        int sum = 0;
        queue<TreeNode*>que;
        que.push(root);
        while(!que.empty()){
            TreeNode* node = que.front();que.pop();
            if(node->left != NULL) {
                que.push(node->left);
                if(node->left->left == NULL && node->left->right == NULL) 
                    sum += node->left->val;
            }
            if(node->right != NULL) {
                que.push(node->right);
            }
        }
        return sum;
    }
};

法二、深搜

左叶子之和 = 左子树的左叶子 + 右子树的左叶子

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

    int sumOfLeftLeaves(TreeNode* root) {
        if(root == NULL) return 0;
        return dfs(root);
    }
   
    
};

注意 sum = 0 一定要放在 dfs内部不能作为全局变量，否则将重复加值

另外一种写法

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