题解 | #二叉树根节点到叶子节点的所有路径和#

层次遍历即可

如果该节点有左儿子，那么更新左儿子的val

如果该节点有右儿子，那么更新右儿子的val

如果该节点是叶子节点，那么res加上该节点的val

 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @return int整型
     */
    int sumNumbers(TreeNode* root) {
        // write code here
        int res=0;
        if(root==NULL) return res;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            int size=q.size();
            while(size--){
                TreeNode* father=q.front();
                q.pop();
                if(father->left==NULL&&father->right==NULL) res+=father->val;
                if(father->left) father->left->val+=10*father->val,q.push(father->left);
                if(father->right) father->right->val+=10*father->val,q.push(father->right);
            }
        }
        return res;
    }
};