Invert Binary Tree.java

E
1525668228
tags: Tree, DFS, BFS

#### DFS
- 简单处理swap
- recursively swap children

#### BFS
- BFS with Queue
- 每次process一个node, swap children; 然后把child加进queue里面
- 直到queue process完

```
/*
Invert a binary tree.

Example
  1         1
 / \       / \
2   3  => 3   2
   /       \
  4         4
Challenge
Do it in recursion is acceptable, can you do it without recursion?

Tags Expand 
Binary Tree

*/


/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
*/


/*
Thoughts:
Use a queue to keep track of nodes. Keep swapping until the queue is processed.
*/
public class Solution {
    public void invertBinaryTree(TreeNode root) {
        if (root == null) {
            return;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()) {
            TreeNode node = queue.poll();
            TreeNode temp = node.left;
            node.left = node.right;
            node.right = temp;
            if (node.left != null) {
                queue.offer(node.left);
            }
            if (node.right != null) {
                queue.offer(node.right);
            }
        }
    }
}


/*
Thoughts: swap every left && right
*/
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        invertTree(root.left);
        invertTree(root.right);
        
        return root;
    }
}



```