226. 翻转二叉树(递归、迭代)

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#二叉树 #leetcode #stack

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本文探讨了两种方法来反转二叉树：递归和迭代。递归采用前序遍历策略，而迭代则通过栈操作避免重复翻转。两种方法对比，展示了编程的不同技巧应用。

题目地址

递归

/**
 * 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) return root;
        swap(root->left, root->right);
        invertTree(root->left);
        invertTree(root->right);
        return root;
    }
};

迭代

/**
 * 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) return root;
        stack<TreeNode*> st;
        st.push(root);
        while (!st.empty()) {
            TreeNode* node;
            node = st.top();
            st.pop();
            if (node) {
                //st.pop();
                if (node->right) st.push(node->right);
                if (node->left) st.push(node->left);
                st.push(node);
                st.push(NULL);
            }
            else {
                // st.pop();
                node = st.top();
                st.pop();
                swap(node->left, node->right);
            }
        }
        return root;
    }
};