LeetCode 226. Invert Binary Tree

Invert a binary tree.

Example:

Input:

     4
   /   \
  2     7
 / \   / \
1   3 6   9

Output:

     4
   /   \
  7     2
 / \   / \
9   6 3   1

解析：翻转二叉树

1.递归的解法

/**
 * 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 == NULL)
            return NULL;
        TreeNode *temp = root->left;
        root->left = invertTree(root->right);
        root->right = invertTree(temp);
        return root;
    }
};

2.非递归的方法，采用层次遍历

/**
 * 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 == NULL)
            return NULL;
        queue<TreeNode*> que;
        que.push(root);
        while(!que.empty())
        {
            TreeNode *node = que.front();
            que.pop();
            TreeNode *temp = node->left;
            node->left = node->right;
            node->right = temp;
            if(node->left)
                que.push(node->left);
            if(node->right)
                que.push(node->right);
        }
        return root;
    }
};