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;
}
原文:http://www.cnblogs.com/juandx/p/6553220.html