解題思路
使用二叉樹的按層遍歷法。
1、將入參節點放入鏈表。
2、判斷鏈表不爲空,則處理當層節點,將當層節點的所有子節點按照從右節點的順序依次放入新的鏈表中
3、將當前鏈表的第一個元素的值放入到返回集合中,
4、將當前鏈表賦值給遍歷鏈表,繼續下一輪的遍歷,直至沒有子節點即可
代碼
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> result = new ArrayList<>();
//root爲Null 則直接返回
if (root == null) {
return result;
}
LinkedList<TreeNode> nodeList = new LinkedList();
LinkedList<TreeNode> childList = null;
nodeList.add(root);
result.add(root.val);
//依次按層遍歷二叉樹
while(nodeList.size() > 0){
childList = new LinkedList();
TreeNode node = null;
while(nodeList.size() > 0){
node = nodeList.removeFirst();
if (node.right != null) {
childList.add(node.right);
}
if (node.left != null) {
childList.add(node.left);
}
}
if (childList.size() > 0) {
result.add(childList.getFirst().val);
}
nodeList = childList;
}
return result;
}
}
作者:zhuan-ye-ban-zhuan
鏈接:https://leetcode-cn.com/problems/binary-tree-right-side-view/solution/java-er-cha-shu-de-you-shi-tu-jie-fa-by-zhuan-ye-b/
來源:力扣(LeetCode)