LeetCode 226 Invert Binary Tree(轉換二叉樹)
阿新 • • 發佈:2017-07-15
public pretty mar containe ret clas move 出錯 uil
翻譯
將下圖中上面的二叉樹轉換為以下的形式。詳細為每一個左孩子節點和右孩子節點互換位置。
原文
如上圖
分析
每次關於樹的題目出錯都在於邊界條件上……所以這次細致多想了一遍:
void swapNode(TreeNode* tree) {
if (tree == NULL || (tree->left == NULL && tree->right == NULL)) {}
else if (tree->left == NULL && tree->right != NULL) {
TreeNode* temp = tree->right;
tree->left = temp;
tree->right = nullptr;
}
else if (tree->right == NULL && tree->left != NULL) {
TreeNode* temp = tree->left;
tree->right = temp;
tree->left = nullptr;
}
else {
TreeNode* temp = tree->left;
tree->left = tree->right;
tree->right = temp;
}
}
不過這樣還不夠,它不過互換了一次。所以我們要用到遞歸:
if(tree->left != NULL)
swapNode(tree->left);
if(tree->right != NULL)
swapNode(tree->right);
最後在題目給定的函數內部調用自己寫的這個遞歸函數就好。
TreeNode* invertTree(TreeNode* root) {
if (root == NULL) return NULL;
swapNode(root);
return root;
}
代碼
/**
* 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:
void swapNode(TreeNode* tree) {
if (tree == NULL || (tree->left == NULL && tree->right == NULL)) {}
else if (tree->left == NULL && tree->right != NULL) {
TreeNode* temp = tree->right;
tree->left = temp;
tree->right = nullptr;
}
else if (tree->right == NULL && tree->left != NULL) {
TreeNode* temp = tree->left;
tree->right = temp;
tree->left = nullptr;
}
else {
TreeNode* temp = tree->left;
tree->left = tree->right;
tree->right = temp;
}
if (tree->left != NULL)
swapNode(tree->left);
if (tree->right != NULL)
swapNode(tree->right);
}
TreeNode* invertTree(TreeNode* root) {
if (root == NULL) return NULL;
swapNode(root);
return root;
}
};
學習
自己的解決方式還是太0基礎,所以來學習學習大神的解法:
TreeNode* invertTree(TreeNode* root) {
if(nullptr == root) return root;
queue<TreeNode*> myQueue; // our queue to do BFS
myQueue.push(root); // push very first item - root
while(!myQueue.empty()){ // run until there are nodes in the queue
TreeNode *node = myQueue.front(); // get element from queue
myQueue.pop(); // remove element from queue
if(node->left != nullptr){ // add left kid to the queue if it exists
myQueue.push(node->left);
}
if(node->right != nullptr){ // add right kid
myQueue.push(node->right);
}
// invert left and right pointers
TreeNode* tmp = node->left;
node->left = node->right;
node->right = tmp;
}
return root;
}
爭取以後少用遞歸了。加油!
LeetCode 226 Invert Binary Tree(轉換二叉樹)