LeetCode.226. 翻转二叉树
难度:easy
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BFS和DFS两种方法:?
/**
* 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;
* }
* }
*/
// DFS
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return root;
}
TreeNode tmpNode = root.left;
root.left = root.right;
root.right = tmpNode;
invertTree(root.left);
invertTree(root.right);
return root;
}
}
//BFS
class Solution {
public TreeNode invertTree(TreeNode root) {
List<Double> ansList = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<>();
if (root == null) {
return null;
}
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode tmpNode = queue.poll();
// 反转左右儿子
TreeNode node = tmpNode.left;
tmpNode.left = tmpNode.right;
tmpNode.right = node;
if (tmpNode.left != null) {
queue.offer(tmpNode.left);
}
if (tmpNode.right != null) {
queue.offer(tmpNode.right);
}
}
}
return root;
}
}
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