LeetCode 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.

For example:
Given the following binary tree,

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

You should return [1, 3, 4].

思路，左右子树的高度，两边一直比较，当前高度更高的，就是从右侧可以看到的。同等高度下，优先看到右子树的。

public List<Integer> rightSideView(TreeNode root) {List<Integer> results = new ArrayList<>();if (root != null) helper(root, 1, 0, results);return results;}private int helper(TreeNode root, int cur, int max, List<Integer> results) {if (cur > max) {results.add(root.val);max = cur;}if (root.right != null) max = helper(root.right, cur + 1, max, results);//如果右子树当前节点更高，更新maxif (root.left != null) max = helper(root.left, cur + 1, max, results);//如果左子树，当前节点更高，更新maxreturn max;}

方法二：

public List<Integer> rightSideView(TreeNode root) {List<Integer> result = new ArrayList<Integer>();if (root == null) return result;result.add(root.val);List<Integer> rsvl = rightSideView(root.left);List<Integer> rsvr = rightSideView(root.right);result.addAll(rsvr);if (rsvl.size() > rsvr.size()) {for (int i = rsvr.size(); i < rsvl.size(); i++)result.add(rsvl.get(i));}return result;}

方法三

 List<Integer> ans = new ArrayList<>();        Deque<TreeNode> q = new ArrayDeque<>();        if (root!=null) { q.addLast(root); }        int curSize = 1, nextSize = 0;        for (curSize=1; !q.isEmpty(); curSize=nextSize, nextSize=0) {            for (int i=0; i<curSize; ++i) {  // there are curSize nodes of current level to be dequeued                TreeNode head = q.removeFirst();                if (i == curSize-1) { ans.add(head.val); }  // add the value of the last node of current level to answer list                if (head.left != null) { q.addLast(head.left); ++nextSize; }                if (head.right != null) { q.addLast(head.right); ++nextSize; }            }        }        return ans;