Leetcode-199: Binary Tree Right Side View

[zeqing-guo]

Description

iven 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].

My Solution

代码的run time是3ms (10.90%)，时间复杂度，空间复杂度。

/**
 * 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) {
        LinkedList<Integer> li = new LinkedList<>();
        LinkedList<TreeNode> lt = new LinkedList<>();
        if (root == null) {
            return li;
        }

        lt.add(root);
        int number = 1;
        while (!lt.isEmpty()) {
            TreeNode element = lt.peek();
            li.add(element.val);
            int nextNumber = 0;
            for (; number > 0; --number) {
                element = lt.poll();
                nextNumber += addChild(lt, element);
            }
            number = nextNumber;
        }
        return li;
    }

    private int addChild(LinkedList<TreeNode> lt, TreeNode element) {
        int number = 0;
        if (element.right != null) {
            lt.add(element.right);
            ++number;
        }
        if (element.left != null) {
            lt.add(element.left);
            ++number;
        }
        return number;
    }
}

Analysis

这个就是一个简单的BFS，没啥好说的。