[数据结构与算法]剑指 Offer II 047. 二叉树剪枝

题目链接：力扣?

思路：递归

递归函数声明：

 TreeNode* pruneTree(TreeNode* root) ;

操作左子树，操作右子树，最后加上根节点(二叉树的后序遍历)

需要把根节点的操作放在最后

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* pruneTree(TreeNode* root) {
if(root==NULL)
return NULL;
TreeNode *l=pruneTree(root->left);
TreeNode *r=pruneTree(root->right);
root->left=l;
root->right=r;
if(!root->left&&!root->right&&root->val==0)
{
    return NULL;
}

return root;
    }
};