1、題目描述
翻轉一棵二叉樹。
2、示例
輸入：
4
/ \
2 7
/ \ / \
1 3 6 9
輸出：
4
/ \
7 2
/ \ / \
9 6 3 1
3、題解
解法一：
基本思想：遞歸
解法二：
基本思想：迭代
#include<iostream>
#include<vector>
#include<algorithm>
#include<deque>
using namespace std;
struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
#define inf 9999
void Init_TreeNode(TreeNode** T, vector<int>& vec, int& pos)
{
if (vec[pos] == inf || vec.size() == 0)
*T = NULL;
else
{
(*T) = new TreeNode(0);
(*T)->val = vec[pos];
Init_TreeNode(&(*T)->left, vec, ++pos);
Init_TreeNode(&(*T)->right, vec, ++pos);
}
}
void PreOrderTraverse(TreeNode* T)
{
if (T == NULL)
return;
cout << T->val << " ";
PreOrderTraverse(T->left);
PreOrderTraverse(T->right);
}
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
//基本思想：遞歸
Recursion(root);
return root;
}
void Recursion(TreeNode* root)
{
if (root == nullptr)
return;
TreeNode* temp = root->left;
root->left = root->right;
root->right = temp;
Recursion(root->left);
Recursion(root->right);
}
};
class Solution1 {
public:
TreeNode* invertTree(TreeNode* root) {
//基本思想：迭代
if (root == nullptr)
return nullptr;
deque<TreeNode*> queue;
queue.push_front(root);
while (!queue.empty())
{
TreeNode* cur = queue.back();
queue.pop_back();
TreeNode* temp = cur->left;
cur->left = cur->right;
cur->right = temp;
if (cur->left != nullptr)
queue.push_front(cur->left);
if (cur->right != nullptr)
queue.push_front(cur->right);
}
return root;
}
};
int main()
{
Solution1 solute;
TreeNode* root = NULL;
vector<int> vec = { 1,2,4,inf,inf,5,inf,inf,3,6,inf,inf,inf };
int pos = 0;
Init_TreeNode(&root, vec, pos);
root = solute.invertTree(root);
PreOrderTraverse(root);
return 0;
}