标签：efi solution rgba str nod output root nbsp 中序
仅供自己学习
题目：
Invert a binary tree.
Example:
Input:
4
/ \
2 7
/ \ / \
1 3 6 9
Output:
4
/ \
7 2
/ \ / \
9 6 3 1
思路：
这就是直接交换数据就可以了，可以前序遍历，后序遍历，中序遍历的交换
代码：
前序遍历：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode() : val(0), left(nullptr), right(nullptr) {} 8 * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} 9 * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} 10 * }; 11 */ 12 class Solution { 13 public: 14 TreeNode* invertTree(TreeNode* root) { 15 if(root == NULL) return NULL; 16 TreeNode* temp; 17 temp=root->left; 18 root->left =root->right; 19 root->right = temp; 20 21 invertTree(root->left); 22 invertTree(root->right); 23 return root; 24 } 25 };
中序遍历：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode() : val(0), left(nullptr), right(nullptr) {} 8 * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} 9 * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} 10 * }; 11 */ 12 class Solution { 13 public: 14 TreeNode* invertTree(TreeNode* root) { 15 if(root == NULL) return NULL; 16 TreeNode* temp; 17 invertTree(root->left); 18 temp=root->left; 19 root->left =root->right; 20 root->right = temp; 21 22 invertTree(root->left); 23 24 return root; 25 } 26 };
后序遍历：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode() : val(0), left(nullptr), right(nullptr) {} 8 * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} 9 * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} 10 * }; 11 */ 12 class Solution { 13 public: 14 TreeNode* invertTree(TreeNode* root) { 15 if(root == NULL) return NULL; 16 TreeNode* temp; 17 invertTree(root->left); 18 invertTree(root->right); 19 temp=root->left; 20 root->left =root->right; 21 root->right = temp; 22 23 24 return root; 25 } 26 };
在leetcode跑，后序遍历时间消耗最佳，前序和中序差别不大
标签：efi solution rgba str nod output root nbsp 中序
原文地址：https://www.cnblogs.com/Mrsdwang/p/14366253.html
