二叉树镜像

方法1:递归实现

先把根节点的左右子树交换,再对左子树、右子树进行同样的操作。

class Solution {
public:
    void Mirror(TreeNode *pRoot) {
        if(pRoot == NULL)
            return;
        TreeNode *temp;
        temp = pRoot->left;
        pRoot->left = pRoot->right;
        pRoot->right = temp;
        Mirror(pRoot->left);
        Mirror(pRoot->right);
    }
};

方法2:非递归实现

既可以用队列辅助,也可以用栈辅助。如果用队列,先把根节点的左右子树交换,然后把左右子树入队列,每次取出队首元素,交换左右子树,直到队列为空。

class Solution {
public:
    void Mirror(TreeNode *pRoot) {
        if(pRoot == NULL)
            return;
        queue<TreeNode*> qq;
        qq.push(pRoot);
        TreeNode *node, *temp;
        while(!qq.empty()){
            node = qq.front();
            temp = node->left;
            node->left = node->right;
            node->right = temp;
            qq.pop();
            if(node->left)
                qq.push(node->left);
            if(node->right)
                qq.push(node->right);
        }
    }
};

如果用栈,则先把根节点的左右子树交换,然后把左右子树分别入栈,每次取栈顶元素,交换左右子树,直到栈为空。

class Solution {
public:
    void Mirror(TreeNode *pRoot) {
        if(pRoot == NULL)
            return;
        stack<TreeNode*> ss;
        TreeNode *node, *temp;
        ss.push(pRoot);
        while(!ss.empty()){
            node = ss.top();
            ss.pop();
            temp = node->left;
            node->left = node->right;
            node->right = temp;
            if(node->left)
                ss.push(node->left);
            if(node->right)
                ss.push(node->right);
        }
    }
};
原文地址:https://www.cnblogs.com/jiaxblog/p/9652303.html