leetcode----199. Binary Tree Right Side View

链接：

https://leetcode.com/problems/binary-tree-right-side-view/

大意：

给定一棵树的根节点root，假设你站在树的最右边，要求找出你所能看到的节点（从上往下）。例子：

思路：

层次遍历，取每层的最右节点即可。

代码：

/**
 * 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;
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        TreeNode ll = root, l = root;
        while (!queue.isEmpty()) {
            TreeNode node = queue.pollFirst();
            if (node.left != null) {
                l = node.left;
                queue.addLast(l);
            }
            if (node.right != null) {
                l = node.right;
                queue.addLast(l);
            }
            if (node == ll) {
                res.add(ll.val); // 层次遍历 取每层的最右节点
                ll = l;
            }
        }
        return res;
    }
}

结果：

结论：

基础题，但得想想最优解。  

改进：（递归版）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int deepestLayer = -1; // 已经走到的最深层
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        myOrder(root, res, 0);
        return res;
    }
    public void myOrder(TreeNode root, List<Integer> res, int currLayer) {
        if (root == null)
            return ;
        if (currLayer > deepestLayer) {
            deepestLayer = currLayer; // 每当走到一个比deepestLayer还深的层时  更新deepestLayer
            res.add(root.val);
        }
        myOrder(root.right, res, currLayer + 1);
        myOrder(root.left, res, currLayer + 1);
    }
}