翻转二叉树-力扣

翻转二叉树，通过前序遍历的顺序，从根节点开始，将节点的左右子节点一次进行交换即可。

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:TreeNode* invertTree(TreeNode* root) {if(root == nullptr){return root;}TreeNode* temp = root->left;root->left = root->right;root->right = temp;invertTree(root->left);invertTree(root->right);return root;}
};

使用迭代的统一写法，前序遍历代码如下：

class Solution {
public:TreeNode* invertTree(TreeNode* root) {stack<TreeNode*> st;if(root != nullptr){st.push(root);}while(!st.empty()){TreeNode* cur = st.top();st.pop();if(cur != nullptr){//st.pop();if(cur->right != nullptr) {st.push(cur->right);}if(cur->left != nullptr) {st.push(cur->left);}st.push(cur);st.push(nullptr);} else {//st.pop();TreeNode * cur = st.top();st.pop();swap(cur->left, cur->right);}}return root;}
};