在 C++ 中查找二叉树克隆中的对应节点
c++server side programmingprogramming
假设我们有两个二叉树，一个是原始二叉树，另一个是克隆二叉树，并给出了对原始树中节点目标的引用。克隆树实际上是原始树的副本。我们必须在克隆树中找到对同一节点的引用。
因此，如果树如下所示，目标是 3，则输出将为 3。
为了解决这个问题，我们将遵循以下步骤 −
定义一个名为solve()的方法，它将获取node1m node2和target
如果node1为null，则返回null
如果node1是target并且node 1的值是node2的值，则返回node2
leftPart := resolve(left of node1, left of node2, target)
rightPart := resolve(right of node1, right of node2, target)
如果leftPart不为null，则返回leftPart，否则返回rightPart
从主方法调用返回solve(original, cloned, target)
示例 (C++)
让我们看下面的实现，以便更好地理解 −
#include <bits/stdc++.h>
using namespace std;
class TreeNode{
public:
int val;
TreeNode *left, *right;
TreeNode(int data){
val = data;
left = 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){
if(val != NULL)
temp->left = new TreeNode(val);
else
temp->left = new TreeNode(0);
return;
} else {
q.push(temp->left);
}
if(!temp->right){
if(val != NULL)
temp->right = new TreeNode(val);
else
temp->right = new TreeNode(0);
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;
}
class Solution {
public:
TreeNode* solve(TreeNode* node1, TreeNode* node2, TreeNode*
target){
if(!node1) return NULL;
if(node1 == target && node1->val == node2->val) return node2;
TreeNode* leftPart = solve(node1->left, node2->left, target);
TreeNode* rightPart = solve(node1->right, node2->right, target);
return leftPart? leftPart : rightPart;
}
TreeNode* getTargetCopy(TreeNode* original, TreeNode* cloned,
TreeNode* target) {
return solve(original, cloned, target);
}
};
main(){
vector<int> v = {7,4,3,NULL,NULL,6,19};
TreeNode *root = make_tree(v);
TreeNode *cloned = make_tree(v);
TreeNode *target = root->right; //The node with value 3
Solution ob;
cout << (ob.getTargetCopy(root, cloned, target))->val;
}
输入
[7,4,3,null,null,6,19] 3
输出
3
