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  • 199. Binary Tree Right Side View

    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,null,4]
    Output: [1, 3, 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, curLevel + 1, list);
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/10201742.html
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