C++ 中的二叉树修剪
c++server side programmingprogramming更新于 2024/9/1 5:14:00
假设我们有一个二叉树的头节点 root，其中每个节点的值都是 0 或 1。我们必须找到相同的树，其中每个不包含 1 的子树都已被删除。因此，如果树像 −
为了解决这个问题，我们将遵循以下步骤 −
定义一个递归方法solve()，它将获取节点。该方法将类似于 −
如果节点为空，则返回空
left of node := solve(left of node)
right of node := solve(right of node)
如果节点左侧为空且节点右侧也为空且节点值为 0，则返回 null
return node
让我们看看下面的实现以便更好地理解 −
示例
#include <bits/stdc++.h>
using namespace std;
class TreeNode{
public:
int val;
TreeNode *left, *right;
TreeNode(int data){
val = data;
left = NULL;
right = NULL;
}
};
void insert(TreeNode **root, int val){
queue<TreeNode*> q;
q.push(*root);
while(q.size()){
TreeNode *temp = q.front();
q.pop();
if(!temp->left){
temp->left = new TreeNode(val);
return;
}else{
q.push(temp->left);
}
if(!temp->right){
temp->right = new TreeNode(val);
return;
}else{
q.push(temp->right);
}
}
}
TreeNode *make_tree(vector<int> v){
TreeNode *root = new TreeNode(v[0]);
for(int i = 1; i<v.size(); i++){
insert(&root, v[i]);
}
return root;
}
void tree_level_trav(TreeNode*root){
if (root == NULL) return;
cout << "[";
queue<TreeNode *> q;
TreeNode *curr;
q.push(root);
q.push(NULL);
while (q.size() > 1) {
curr = q.front();
q.pop();
if (curr == NULL){
q.push(NULL);
} else {
if(curr->left)
q.push(curr->left);
if(curr->right)
q.push(curr->right);
if(curr == NULL){
cout << "null" << ", ";
}else{
cout << curr->val << ", ";
}
}
}
cout << "]"<<endl;
}
class Solution {
public:
TreeNode* pruneTree(TreeNode* node) {
if(!node)return NULL;
node->left = pruneTree(node->left);
node->right = pruneTree(node->right);
if(!node->left && !node->right && !node->val){
return NULL;
}
return node;
}
};
main(){
vector<int> v = {1,1,0,1,1,0,1,0};
TreeNode *root = make_tree(v);
Solution ob;
tree_level_trav(ob.pruneTree(root));
}
输入
[1,1,0,1,1,0,1,0]
输出
[1, 1, 0, 1, 1, 1, ]
