leetcode:Invert Binary Tree

Invert a binary tree.

     4
   /     2     7
 / \   / 1   3 6   9

to

     4
   /     7     2
 / \   / 9   6 3   1即反转二叉树,代码如下:
/**
 * 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 {        //在每一层将节点的左孩子和右孩子分别交换即可,直到节点没有孩子(递归recursion)
public:
    TreeNode* invertTree(TreeNode* root) {
        TreeNode *temp;
        if(root)
        {
            temp = root->right;
            root->right = invertTree(root->left);
            root->left  = invertTree(temp);
        }
        return root;
    }
};

  或者:

class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (root) {
            root->left = invertTree(root->left), root->right = invertTree(root->right);
            std::swap(root->left, root->right);
        }
        return root;
    }
};

  其他解法:

1、C++ using queue, 0ms

非递归版本,做广度优先处理,使用一个队列来保存当前水平未处理的非空节点。每次,切换前节点的左,右指针,将其弹出,并添加它的非空孩子到队列中。

class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
    if (root==nullptr) return nullptr;

    queue<TreeNode*> st;
    st.push(root);

    while(!st.empty()){
        TreeNode* n=st.front();
        st.pop();

        if (n->left!=NULL)
            st.push(n->left);
        if (n->right!=NULL)
            st.push(n->right);
        swap(n->left,n->right);

    }
    return root;
}
};

  

2、先前序遍历这棵树的每个结点,如果遍历到的结点有子节点,就交换他的两个子节点,当交换完所有非叶子节点的左右子节点之后,就得到了该二叉树的镜像。

class Solution {
public:
     TreeNode* invertTree(TreeNode* root) {
         if(root==null){
            return null;
        }
        if(root->left==null&&root->right==null){
            return root;
        }
        TreeNode *temp = root->left;
        root->left = root->right;
        root->right = temp;
        if(root->left!=null){
            invertTree(root->left);
        }
        if(root->right!=null){
            invertTree(root->right);
        }
        return root;
    }
};

  显示出错:‘null’ was not declared in this scope

将null改为nullptr就 Accepted了:

class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
         if(root==nullptr){
            return nullptr;
        }
        if(root->left==nullptr&&root->right==nullptr){
            return root;
        }
        TreeNode *temp = root->left;
        root->left = root->right;
        root->right = temp;
        if(root->left!=nullptr){
            invertTree(root->left);
        }
        if(root->right!=nullptr){
            invertTree(root->right);
        }
        return root;
    }
};

【C++11】nullptr关键字:从1972年C语言刚刚诞生以来,常数0就扮演着整数(int)0和空指针( null pointer )两种角色。为了避免理解上的二义性,C语言通常使用NULL宏来表示空指针,NULL宏通常被定义为(void *)0或0, 而C++仅仅采用0来表示空指针,这样存在一个问题:比如对于重载函数 fun(char *) 和 fun(int) 的调用来说,若直接用NULL作为参数调用fun(NULL),我们可能认为NULL作为空指针的表示,应该调用 fun(char *) 函数,然而NULL实际上的值为0,就会调用  fun(int) 。 nullptr 关键字正是为了解决这一问题而产生的。

nullptr 能够转换成任何指针类型(包括成员函数指针和成员变量指针)和bool类型(这是为了兼容普通指针都能使用 if(ptr) 判断是否为空指针的形式),但是不能被转换为整数0。 

(更多了解可参考:http://blog.csdn.net/huang_xw/article/details/8764346)

时间: 2024-11-08 19:43:03

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