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].
一刷
題解：用遞歸來求解滓侍，里面最巧妙的是惊奇，遞歸子問題內(nèi)傳入了level， 如果level與我們當(dāng)前的array的size想同喉刘，那么它就不是rightmost
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> res = new ArrayList<>();
rightView(root, 0, res);
return res;
}
private void rightView(TreeNode root, int level, List<Integer> res){
if(root==null) return;
if(level == res.size()) res.add(root.val);
rightView(root.right, level+1, res);
rightView(root.left, level+1, res);
}
}
二刷
其實(shí)就是DFS庐氮，但是要加上一個(gè)level的信息语稠，來決定是否加入結(jié)果中。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> res = new ArrayList<>();
if(root == null) return res;
rightView(root, 1, res);
return res;
}
private void rightView(TreeNode root, int level, List<Integer> res){
if(root == null) return;
if(level>res.size()) res.add(root.val);
rightView(root.right, level+1, res);
rightView(root.left, level+1, res);
}
}
三刷
DFS + level信息
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> res = new ArrayList<>();
rightSideView(root, 1, res);
return res;
}
private void rightSideView(TreeNode root, int level, List<Integer> res){
if(root == null) return;
if(level>res.size()){
res.add(root.val);
}
rightSideView(root.right, level+1, res);
rightSideView(root.left, level+1, res);
}
}
