［LeetCode］Invert Binary Tree

解题思路：

递归法，只要root有一个子树不为空，就要swap一次，然后深入到子树中执行相同的操作

前条件：存在一个root

不变式：swap左右子树，递归

结束条件：root左右子树都为NULL

临界条件：root为NULL

// 编译错误

Line 23: return-statement with a value, in function returning 'void' [-fpermissive]

返回类型为void，用return NULL是肯定不对的 

/** * 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;        }        swapNode(root);        return root;    }    void swapNode(TreeNode* root){        if (root->left == NULL && root->right == NULL){            return;        }else{            TreeNode* temp;            temp = root->left;            root->left = root->right;            root->right = temp;            if (root->left != NULL){                invertTree(root->left);            }            if (root->right != NULL){                invertTree(root->right);            }        }    }};