第27题:Leetcode226: Invert Binary Tree反转二叉树

翻转一棵二叉树。

示例:

输入:

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

输出:

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

 思路

如果根节点存在,就交换两个子树的根节点,用递归,从下至上。、

解一:

auto tmp = invertTree(root->left);

将左子树整体 当作一个节点 交换。

最后返回根节点。

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

解二: 

只单单交换左右两个节点。

如果交换完后,左子树存在,就交换左子树,右子树存在,就交换右子树

/**
 * 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 root;
        TreeNode *tmp = root->left;
        root->left = root->right;
        root->right = tmp;
        if(root->left)invertTree(root->left);
        if(root->right)invertTree(root->right);
        return root;
        
    }
};

 


解三:

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    void Mirror(TreeNode *pRoot) {
        
        if(!pRoot)
            return;
        if(!pRoot->left&&!pRoot->right)
            return;
        
        auto temp=pRoot->left;
        pRoot->left=pRoot->right;
        pRoot->right=temp;
        
        if(pRoot->left)
            Mirror(pRoot->left);
        if(pRoot->right)
             Mirror(pRoot->right);

    }
};

 

posted @ 2019-02-07 19:26 lightmare 阅读(...) 评论(...) 编辑 收藏