题目

invert-binary-tree

2. 算法

* 递归

* 非递归

3. 代码

* 递归

// Recursion
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (!root) return NULL;
        TreeNode *tmp = root->left;
        root->left = invertTree(root->right);
        root->right = invertTree(tmp);
        return root;
    }
};

* 非递归

// Non-Recursion
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (!root) return NULL;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            TreeNode *node = q.front(); q.pop();
            TreeNode *tmp = node->left;
            node->left = node->right;
            node->right = tmp;
            if (node->left) q.push(node->left);
            if (node->right) q.push(node->right);
        }
        return root;
    }
};