Binary Tree Right Side View:查找二叉树最右支
2018-01-10 19:44
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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,
You should return
思路:可以广度优先,也可以深度优先。我选择的是深度优先,因为写着简单。
中-右-左 + 深度,就可以解决这个问题。不是很难。
时空复杂度均为O(N)。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public void dfs(TreeNode n,List<Integer> l,int depth){
if(n==null) return;
if(l.size() < depth) l.add(n.val);
dfs(n.right,l,depth+1);
dfs(n.left,l,depth+1);
}
public List<Integer> rightSideView(TreeNode root) {
List<Integer> RES = new ArrayList<Integer>();
dfs(root ,RES ,1);
return RES;
}
}
For example:
Given the following binary tree,
1 <--- / \ 2 3 <--- \ \ 5 4 <---
You should return
[1, 3, 4].
思路:可以广度优先,也可以深度优先。我选择的是深度优先,因为写着简单。
中-右-左 + 深度,就可以解决这个问题。不是很难。
时空复杂度均为O(N)。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public void dfs(TreeNode n,List<Integer> l,int depth){
if(n==null) return;
if(l.size() < depth) l.add(n.val);
dfs(n.right,l,depth+1);
dfs(n.left,l,depth+1);
}
public List<Integer> rightSideView(TreeNode root) {
List<Integer> RES = new ArrayList<Integer>();
dfs(root ,RES ,1);
return RES;
}
}
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