[LeetCode]199. Binary Tree Right Side View

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

返回从右边看看到的数组值的list

我的解法：

把左右子树从右边看的结果返回，然后把左子树中index大于右子树长度的部分加进去，然后加上root的值，得到结果。

public class Solution {    public List<Integer> rightSideView(TreeNode root) {        if (root == null) {            return new LinkedList();        }        List<Integer> left = rightSideView(root.left);        List<Integer> right = rightSideView(root.right);        for (int i = right.size(); i < left.size(); i++) {            right.add(left.get(i));        }        right.add(0, root.val);        return right;    }}

LeetCode解法：

当前depth等于list时候，然后先遍历右子树，在遍历左子树，把当前节点值加入到list中

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