【LeetCode】199. Binary Tree Right Side View

Problem：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.

题目：给定一个二叉树，输出从右侧看到的节点（大致就是每层的最右侧的节点）.

思路：做的二叉树的深度遍历

1.节点是否为空，不为空则记录节点。

2.递归查找根节点的左孩子，是否为空，不为空记录在该层节点数组中

3.递归查找根节点的右孩子，是否为空，不为空记录在该层节点数组中

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
	public List<Integer> rightSideView(TreeNode root) {
		List<Integer> ret = new ArrayList<Integer>();
		if (root == null) {
			return ret;
		}
		int height = getHeight(root, 0);
		TreeNode[] edgeMap = new TreeNode[height];
		setRightEdge(root, 0, edgeMap);
		for (int i = 0; i < height; i++) {
			ret.add(edgeMap[i].val);
		}
		return ret;
	}
	// refined for this problem
	public void setRightEdge(TreeNode node, int l, TreeNode[] rightEdge) {
		if (node == null) {
			return;
		}
		rightEdge[l]=node;
		setRightEdge(node.left, l+1,rightEdge);
		setRightEdge(node.right, l+1,rightEdge);
	}
	// get the height of treenode 
	public int getHeight(TreeNode node, int l) {
		if (node == null) {
			return l;
		}
		return Math.max(getHeight(node.left, l + 1), getHeight(node.right, l + 1));
	}
}