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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;
}
}
```