继续刷LeetCode,第226题,给定一个输入二叉树,求交换后的二叉树。
分析:
方法一:交换二叉树,就是通过递归的方式,从上到下遍历树的每个节点,并且交换节点的左右孩子,最后返回根节点。
方法二:既然可以从上到下遍历,那么也可以从下岛上遍历,并交换左右孩子。
问题:
1、交换的方式,是交换左右孩子。如果交换左右孩子的值,那么碰到NULL的情况就加大处理的难度。
附上C++方法一的代码:
/**
* 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 NULL;
TreeNode* temp=root->left;
root->left=root->right;
root->right=temp;
invertTree(root->left);
invertTree(root->right);
return root;
}
};
附上方法二的C++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 NULL;
TreeNode* leftNode=root->left;
TreeNode* rightNode=root->right;
if(root->left!=NULL)
invertTree(root->left);
if(root->right!=NULL)
invertTree(root->right);
root->left=rightNode;
root->right=leftNode;
return root;
}
};
附上Python代码1:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
if root==None:
return root
tmp=root.left
root.left=root.right
root.right=tmp
self.invertTree(root.left)
self.invertTree(root.right)
return root
附上简化的Python代码:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
if root==None:
return root
tmp=root.left
root.left=self.invertTree(root.right)
root.right=self.invertTree(tmp)
return root