题目描述
翻转一棵二叉树。
样例描述
输入:
4
/ \
2 7
/ \ / \
1 3 6 9
输出:
4
/ \
7 2
/ \ / \
9 6 3 1
思路
方法一:递归 (也就是DFS)
- 不断地压栈,(根结点要翻转它左右子树,先让左右子树去翻转,左右子树的根结点想翻转也让各自的子树去翻转,一级一级递归下去,直到叶子结点,没有左右子树,往上返回,同时进行翻转)。
方法二:迭代,层序遍历BFS
- 将每一层的结点入队,只要队列不空,每拿出一个结点,就翻转其左右孩子(就算没有都是null,交换也不影响),然后看左右孩子是否为空,不为空就入队。
代码
递归
/**
* 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 {
public TreeNode invertTree(TreeNode root) {
if (root == null) return root;
//先递归让左右子树都交换好
if (root.left != null) invertTree(root.left);
if (root.right != null) invertTree(root.right);
TreeNode t = root.left;
root.left = root.right;
root.right = t;
return root;
}
}
迭代
/**
* 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 {
public TreeNode invertTree(TreeNode root) {
Deque<TreeNode> q = new LinkedList<>();
if (root != null) q.offer(root);
while (!q.isEmpty()) {
TreeNode node = q.poll();
//交换左右子树结点(空也没事,交换不影响)
TreeNode t = node.left;
node.left = node.right;
node.right = t;
//如果左右子树不为空,就加入队列
if (node.left != null) q.offer(node.left);
if (node.right != null) q.offer(node.right);
}
return root;
}
}