标签：binary 推荐 code 思路 tput ret swap pch 层次遍历
Invert a binary tree.
Example:
Input:
4 / 2 7 / \ / 1 3 6 9
Output:
4 / 7 2 / \ / 9 6 3 1
翻转一棵二叉树。EZ题。
方法1：整体思路就是层次遍历，然后把左右孩子对调。
1 class Solution { 2 public: 3 TreeNode* invertTree(TreeNode* root) 4 { 5 queue<TreeNode*> q; 6 if(root!=nullptr) 7 { 8 q.push(root); 9 while(!q.empty()) 10 { 11 TreeNode* node = q.front(); 12 q.pop(); 13 swapChild(node); 14 if(node->left!=nullptr) 15 q.push(node->left); 16 if(node->right!=nullptr) 17 q.push(node->right); 18 } 19 } 20 return root; 21 } 22 void swapChild(TreeNode* node) 23 { 24 if(node != nullptr) 25 { 26 TreeNode* temp = node->right; 27 node->right = node->left; 28 node->left = temp; 29 } 30 } 31 };
方法2：递归求解，简洁的让人觉得可怕,跑的还飞快，推荐食用??
class Solution { public: TreeNode* invertTree(TreeNode* root) { if(root!=nullptr) { auto temp = root->right; root->right = invertTree(root->left); root->left = invertTree(temp); } return root; } };
标签：binary 推荐 code 思路 tput ret swap pch 层次遍历
原文地址：https://www.cnblogs.com/Swetchine/p/11161400.html
