226. Invert Binary Tree

Invert a binary tree.

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

to

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

不用多說了，就是反轉二元樹

方法有好幾種。我這邊用最快的方法BFS的作法來做

/**
 * 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:
  
  TreeNode* invertTree(TreeNode* root) {

    if(nullptr == root) return root;

    stack<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.top();  // 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      
        swap(node->left, node->right);

    }
    return root;
}       
};