Binary Tree Right Side View

題目：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.

解題思路：
1. 右子樹的高度 >= 左子樹 ， easy ;
2. 左子樹 > 右子樹?
3. 通過層序遍歷(BFS)，每一層的最右邊加入結果鏈表；

/**
 * 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 ArrayList<Integer> ();
        LinkedList<TreeNode> queue = new LinkedList<TreeNode> ();
        //當前層節點個數，下一層節點個數
        int curCount , nextCount = 0;
        if(root == null) return list;
        queue.add(root);
        curCount = 1;
        list.add(root.val);

        while(!queue.isEmpty()) {
            if(curCount == 0) {
                curCount = nextCount;
                // 注意這裏一定要置0
                nextCount = 0;
                list.add(queue.getLast().val);
            }
            TreeNode node = queue.removeFirst();
            curCount--;
            if(node.left != null) {
                queue.add(node.left);
                nextCount++;
            }
            if(node.right != null) {
                queue.add(node.right);
                nextCount++;
            }
        }

        return list;
    }
}

DFS(前序遍歷)解題

• 因爲每一層都必然有一個會有一個值添加到返回鏈表中；
• 這樣我們只要比較返回鏈表的大小與當前的深度，如果小於，則添加到鏈表

public class Solution {
        public List<Integer> rightSideView(TreeNode root) {
            List<Integer> result = new ArrayList<Integer>();
            help(root, 1, result);
            return result;
        }

        public void help(TreeNode root, int depth, List<Integer> result) {
            if (root == null)
                return;
            if (result.size() < depth)
                result.add(root.val);
            // 從右子樹開始!
            help(root.right, depth + 1, result);
            help(root.left, depth + 1, result);
        }
    }