226. Invert Binary Tree

Invert a binary tree.

     4
   /   \
  2     7
 / \   / \
1   3 6   9

to

     4
   /   \
  7     2
 / \   / \
9   6 3   1

我的解法：

/**
 * 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 Invert(TreeNode* root){
         if(root){
            TreeNode* temp = root->left;
            root->left = root->right;
            root->right = temp;
            if(root->left)invertTree(root->left);
            if(root->right)invertTree(root->right);
        }else{
            return;
        }
    }
    
    TreeNode* invertTree(TreeNode* root) {
        Invert(root);
        return root;
    }
};

寫的很臃腫，下面式別人的簡潔的

Recursive

TreeNode* invertTree(TreeNode* root) {
    if (root) {
        invertTree(root->left);
        invertTree(root->right);
        std::swap(root->left, root->right);
    }
    return root;
}
下面這個是用棧實現的非迭代的。 

TreeNode* invertTree(TreeNode* root) {
    std::stack<TreeNode*> stk;
    stk.push(root);
    
    while (!stk.empty()) {
        TreeNode* p = stk.top();
        stk.pop();
        if (p) {
            stk.push(p->left);
            stk.push(p->right);
            std::swap(p->left, p->right);
        }
    }
    return root;
}