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].

分析：

这题有迷惑性，不要以为只是让你找出最右边的节点，如果左边节点比右边节点高，那么左边节点的右边部分也要输出来。

这题可以按层获取树的节点，然后把每层最右边的找出来。

/**
 * 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> result = new ArrayList<>();
        if (root == null) return result;

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode top = queue.poll();
                if (i == size - 1) {
                    result.add(top.val);
                }
                if (top.left != null) {
                    queue.add(top.left);
                }
                if (top.right != null) {
                    queue.add(top.right);
                }
            }
        }
        return result;
    }
}

 

public class Solution {
    // The core idea of this algorithm:
    //     1.Each depth of the tree only select one node.
    //     2. View depth is current size of result list.
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        rightView(root, result, 0);
        return result;
    }
    
    public void rightView(TreeNode curr, List<Integer> result, int currDepth){
        if(curr == null) return;
        if(currDepth == result.size()){
            result.add(curr.val);
        }
        rightView(curr.right, result, currDepth + 1);
        rightView(curr.left, result, currDepth + 1);
    }
}