算法:二叉树从右侧看到的内容 199. Binary Tree Right Side View

Posted 2021-12-10 架构师易筋

199. Binary Tree Right Side View

Given the root of 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.

Example 1:

Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]

Example 2:

Input: root = [1,null,3]
Output: [1,3]

Example 3:

Input: root = []
Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

DFS深度优先解法

从右侧看过来，如果只有左侧，因为右侧为空，所以也能看到左侧。

				1
			/
		2
	/
3

比如上面的结果，应该是1，2，3. 所以问题转换为每一层只存一个数，从右节点开始遍历。

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