问题描述
问题链接:https://leetcode.com/problems/invert-binary-tree/#/description
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.
我的代码
看了这个描述觉得Google也是逗啊,哈哈。
/**
* 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) {
/*
思路就是递归的交换左右子节点
*/
if(root == null){
return root;
}
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
invertTree(root.left);
invertTree(root.right);
return root;
}
}
打败了30.77%的Java代码,来看看讨论区。
讨论区
Straightforward DFS recursive, iterative, BFS solutions
链接地址:https://discuss.leetcode.com/topic/16039/straightforward-dfs-recursive-iterative-bfs-solutions
用了栈和BFS,感觉挺新鲜。
As in many other cases this problem has more than one possible solutions:
Lets start with straightforward - recursive DFS - it’s easy to write and pretty much concise.
public class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
final TreeNode left = root.left,
right = root.right;
root.left = invertTree(right);
root.right = invertTree(left);
return root;
}
}
The above solution is correct, but it is also bound to the application stack, which means that it’s no so much scalable - (you can find the problem size that will overflow the stack and crash your application), so more robust solution would be to use stack data structure.
public class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
final Deque<TreeNode> stack = new LinkedList<>();
stack.push(root);
while(!stack.isEmpty()) {
final TreeNode node = stack.pop();
final 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);
}
}
return root;
}
}
Finally we can easly convert the above solution to BFS - or so called level order traversal.
public class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
final Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()) {
final TreeNode node = queue.poll();
final TreeNode left = node.left;
node.left = node.right;
node.right = left;
if(node.left != null) {
queue.offer(node.left);
}
if(node.right != null) {
queue.offer(node.right);
}
}
return root;
}
}
If I can write this code, does it mean I can get job at Google? ;)