Approach #1 (Recursive)1
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38/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
// In place
public TreeNode invertTree(TreeNode root) {
if(root!=null){
TreeNode right = invertTree(root.right);
TreeNode left = invertTree(root.left);
root.left = right;
root.right = left;
return root;
}
return null;
}
/* New nodes
public TreeNode invertTree(TreeNode root) {
if(root!=null){
TreeNode newRoot=new TreeNode(root.val);
newRoot.left=invertTree(root.right);
newRoot.right=invertTree(root.left);
return newRoot;
}
return null;
}*/
}
Approach #2 (Iterative) using queue1
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16class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) return null;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while (!queue.isEmpty()) {
TreeNode current = queue.poll();
TreeNode tmp = current.left;
current.left = current.right;
current.right = tmp;
if (current.left != null) queue.add(current.left);
if (current.right != null) queue.add(current.right);
}
return root;
}
}