Leetcode-404: sum of left leaves

404. Sum of left leaves
Title Description ：
Given the root node of a binary tree root , Returns the sum of all left leaves .

Depth first （O(N),O(N)）

class Solution {
    
public:
    void  getLeftNum(TreeNode* root, int &n){
    
        //  If the root node is empty or there is only one node, it returns 
        if(!root->left && !root->right || root == nullptr )  return;
        //  Add the left leaf of this node 
        else if(root->left != nullptr && root->left->left == nullptr && root->left->right==nullptr) n+=root->left->val;
        // Otherwise, continue to process the left subtree 
        else if(root->left)
            getLeftNum(root->left,n);
        //  Deal with the right subtree  
        if(root->right)
            getLeftNum(root->right,n);
    }
    int sumOfLeftLeaves(TreeNode* root) {
    
        int n =0 ;
        getLeftNum(root,n);
        return n;
    }
};

breadth-first （O(N),O(N)）

class Solution {
    
public:
    int sumOfLeftLeaves(TreeNode* root) {
    
        if(!root->left && !root->right || root==nullptr) return 0;
        //  Initialization return value 
        int res=0;
        //  Breadth first traversal 
        queue<TreeNode*> que;
        que.push(root);
        //  Breadth first traversal processing template 
        while(!que.empty()){
    
            TreeNode* node = que.front();
            que.pop();
            //  The left child joined the team 
            if(node->left){
    
                que.push(node->left) ;
                //  If the left child is a leaf node, add it 
                if(node->left->left == nullptr && node->left->right == nullptr)
                res+=node->left->val;
            }  
            //  The right kid joins the team 
            if(node->right) que.push(node->right);
        }
        return res;    
    }
};