Leetcode 199. Binary Tree Right Side View-优快云博客

本文链接：https://blog.youkuaiyun.com/GoodJobJasper/article/details/112646521

这篇博客介绍了两种方法来获取二叉树的右视图：一种是使用广度优先搜索（BFS）的双队列迭代法，另一种是深度优先搜索（DFS）的根-右-左顺序遍历。这两种方法的时间复杂度均为O(n)，空间复杂度分别为O(n)和O(h)。BFS方法通过逐层遍历并记录每层最后一个节点的值来获取右视图，而DFS方法则沿着特定顺序递归遍历树节点。

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方法1: bfs-two queues-iteration。这道题目肯定是用bfs更加清楚明了。时间复杂n，空间复杂n。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if(root == null) return list;
        Queue<TreeNode> currLevel = new LinkedList<>();
        Queue<TreeNode> nextLevel = new LinkedList<>();
        nextLevel.offer(root);
        while(!nextLevel.isEmpty()){
            currLevel = new LinkedList<>(nextLevel);
            nextLevel.clear();
            list.add(currLevel.peek().val);
            for(TreeNode node : currLevel){
                if(node.right != null) nextLevel.offer(node.right);
                if(node.left != null) nextLevel.offer(node.left);
            }
        }
        return list;
    }
}

方法2: dfs-(root-right-left order)。这道题目我们用root-
right-left的顺序来遍历tree。时间复杂n，空间复杂h。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    List<Integer> list = new ArrayList<>();
    public List<Integer> rightSideView(TreeNode root) {
        if(root == null) return list;
        dfsPreorder(root, 0);
        return list;
    }
    
    public void dfsPreorder(TreeNode root, int level){
        if(list.size() == level) {
            list.add(root.val);
        }
        if(root.right != null) dfsPreorder(root.right, level + 1);
        if(root.left != null) dfsPreorder(root.left, level + 1);
    }
}