本文是对 LeetCode Binary Tree Right Side View 解法的探讨。
题目:
Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
For example:
Given the following binary tree,
1 <---
/ 2 3 <---
\ 5 4 <---
You should return [1, 3, 4].
解法如下:
/**
* 因为题目的二叉树并不是满二叉树,所以采用层序遍历的方式。
* 将以前层序遍历中一个个出队的方式变为一层层出队,
* 这样就能定位最右边的节点。
* @param root
* @return
*/
public List<Integer> rightSideView(TreeNode root) {
List<Integer> list = new ArrayList<>();
if (root == null) {
return list;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
//将每一层的节点都出队
for (int i = 0; i < size; i++) {
TreeNode treeNode = queue.poll();
if (i == 0) {
list.add(treeNode.val);
}
if (treeNode.right != null) {
queue.offer(treeNode.right);
}
if (treeNode.left != null) {
queue.offer(treeNode.left);
}
}
}
return list;
}