05 August 2008
Invert a binary tree.
Example:
Input:
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4
/ \
2 7
/ \ / \
1 3 6 9
Output:
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5
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 f*** off.
同样可以用递归，DFS和BFS来做。
递归的写法很简单，只是复杂度没有什么好优化的。
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {//递归的终止条件
return root;
}
TreeNode left = invertTree(root.left);
TreeNode right = invertTree(root.right);
//递归调用后的最后一步
root.left = right;
root.right = left;
return root;
}
}
DFS，可以用stack或者deque
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return root;
}
Stack<TreeNode> stack = new Stack<>();
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);
}
}
return root;
}
}
BFS，其实就是层序遍历level order traversal，数据结构不一样
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return root;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);//vs add()
while (!queue.isEmpty()) {
TreeNode node = queue.poll();//vs peek()
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);
}
}
return root;
}
}