LeetCode 226 Invert Binary Tree

Problem:

Invert a binary tree.

     4
   /   
  2     7
 /    / 
1   3 6   9

to

     4
   /   
  7     2
 /    / 
9   6 3   1

Analysis:

1、helper function + recursion

一开始我用调用子函数，在子函数中递归的方式实现：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    void swapNode (TreeNode* root) {
        TreeNode *tmp = root->left;
        root->left = root->right;
        root->right = tmp;
        
        if (root->left != NULL) {
            swapNode (root->left);
        }
        
        if (root->right != NULL) {
            swapNode (root->right);
        }
    }
    
public:
    TreeNode* invertTree(TreeNode* root) {
        if (root != NULL) {
            swapNode (root);
        }
        
        return root;
    }
};

虽然解决了问题，但效率太低

2、recursion

进而将递归写入主函数中，并简化了部分代码

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

效率有所提高

3、非递归方法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (root == NULL) {
            return NULL;
        }
        
        queue<TreeNode*> q;
        q.push(root);
        
        while (!q.empty()) {
            TreeNode* cur = q.front();
            q.pop();
            
            TreeNode* tmp = cur->left;
            cur->left = cur->right;
            cur->right = tmp;
            
            if (cur->left != NULL) {
                q.push(cur->left);
            }
            
            if (cur->right != NULL) {
                q.push(cur->right);
            }
        }
        
        return root;
    }
};