Lintcode：翻转二叉树

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问题：

翻转一棵二叉树。左右子树交换。

样例：

样例 1:

输入: {1,3,#}
输出: {1,#,3}
解释:
	  1    1
	 /  =>  \
	3        3

样例 2:

输入: {1,2,3,#,#,4}
输出: {1,3,2,#,4}
解释: 
	
      1         1
     / \       / \
    2   3  => 3   2
       /       \
      4         4

python：

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: a TreeNode, the root of the binary tree
    @return: nothing
    """
    def invertLeftAndRight(self, root):
        if root == None:
            return root
        temp = root.left
        root.left = self.invertLeftAndRight(root.right)
        root.right = self.invertLeftAndRight(temp)
        return root
    def invertBinaryTree(self, root):
        # write your code here
        self.invertLeftAndRight(root)

C++：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    TreeNode *invertLeftAndRight(TreeNode *root)
    {
        if(root == NULL)
        {
            return root;
        }
        TreeNode *temp = root->left;
        root->left = invertLeftAndRight(root->right);
        root->right = invertLeftAndRight(temp);
        return root;
    }
    void invertBinaryTree(TreeNode * root) {
        // write your code here
        root = invertLeftAndRight(root);
    }
};