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;
    }
};