Invert a binary tree.
4
/ \
2 7
/ \ / \
1 3 6 9
to
4
/ \
7 2
/ \ / \
9 6 3 1
Trivia:This problem was inspired by this original tweet by Max Howell:Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.
方法1:recursive
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode invertTree(TreeNode root) {
if(root == null) return null;
TreeNode left = root.left;
TreeNode right = root.right;
root.left = invertTree(right);
root.right = invertTree(left);
return root;
}
}
但是树太深会有stack overflow的风险
方法2: BFS - Stack
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode invertTree(TreeNode root) {
if(root == null) return null;
Deque<TreeNode> stack = new LinkedList<>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode node = stack.pop();
TreeNode left = node.left;
node.left = node.right;
node.right = left;
if(node.left!=null){
stack.push(node.left);
}
if(node.right!=null){
stack.push(node.right);
}
}
}
}
