199. Binary Tree Right Side View - Medium

发布时间：2020-12-14 05:17:36 所属栏目：大数据来源：网络整理

导读：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. Example: Input:?[1,2,3,null,5,4]Output:?[1,4]Explanation: 1 --- / 2 3 --- 5 4 --- ? M1: BFS le

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.

Example:

Input:?[1,2,3,null,5,4]
Output:?[1,4]
Explanation:

   1            <---
 /   2     3         <---
        5     4       <---

M1: BFS

level order traverse,，每次遍历完一层后，只把这一层最右的数字加入res中

time: O(n),space: O(N)　　-- N: most number of nodes in one level

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null) {
            return res;
        }
        
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while(!q.isEmpty()) {
            List<Integer> tmp = new ArrayList<>();
            int size = q.size();
            for(int i = 0; i < size; i++) {
                TreeNode t = q.poll();
                tmp.add(t.val);
                if(t.left != null) {
                    q.offer(t.left);
                }
                if(t.right != null) {
                    q.offer(t.right);
                }
            }
            res.add(tmp.get(tmp.size() - 1));
        }
        
        return res;
    }
}

M2: DFS

ref:?https://leetcode.com/problems/binary-tree-right-side-view/discuss/56012/My-simple-accepted-solution(JAVA)

time: O(n),space: 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 List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null) {
            return res;
        }
        dfs(root,0,res);
        return res;
    }
    
    public void dfs(TreeNode node,int curLevel,List<Integer> list) {
        if(node == null) {
            return;
        }
        if(curLevel == list.size()) {
            list.add(node.val);
        }
        
        dfs(node.right,curLevel + 1,list);
        dfs(node.left,list);
    }
}

（编辑：李大同）

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