尴尬了,昨天被问,突然不知道咋搞。
二叉树遍历,前中后,是以root为准的前中后
所以反转二叉树用后续遍历就好
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
递归
struct TreeNode* invertTree(struct TreeNode* root)
{
struct TreeNode *node;
if (root == NULL)
return root;
node = invertTree(root->left);
root->left = invertTree(root->right);
root->right = node;
return root;
}
非递归
struct TreeNode* invertTree(struct TreeNode *root)
{
struct TreeNode *node, *tmp;
Stack treeStack;
if (root == NULL)
return root;
stack_init(&treeStack, NULL);
stack_push(&treeStack, root);
while (treeStack.size > 0)
{
stack_pop(&treeStack, &node);
tmp = node->left;
node->left = node->right;
node->right = tmp;
if (node->left)
stack_push(&treeStack, node->left);
if (node->right)
stack_push(&treeStack, node->right);
}
stack_destory(&treeStack);
return root;
}