題目描述

給定一個二叉樹的?根節點?root，想象自己站在它的右側，按照從頂部到底部的順序，返回從右側所能看到的節點值。

輸入:?[1,2,3,null,5,null,4] 輸出:?[1,3,4] 輸入:?[1,null,3] 輸出:?[1,3] 輸入:?[] 輸出:?[]

解題思路

思路一：采用兩個隊列的方法，第一個隊列記錄第一層，第二個隊列記錄第二層。知道兩個隊列都為空的時候，遍歷結束。時間復雜度和空間復雜度都是O（N）

但這種方法有一個不太好的地方，就是代碼偏多，需要定義兩個隊列。面試官姐姐可能并不喜歡你的方法。

思路二：采用一個隊列+計數的方法，每個層次的遍歷，之前，都先統計一下這一層有多少個節點。如何統計呢？其實就是當時隊列中元素的個數。例如，隊列有n個節點，當這一層級的第n個節目點出棧的時候，將該節點計入結果。

解題核心代碼

class Solution {public List<Integer> rightSideView(TreeNode root) {LinkedList<Integer> res = new LinkedList<>();if (root==null){return res;}LinkedList<TreeNode> queue = new LinkedList<>();queue.offer(root);while (!queue.isEmpty()){int size = queue.size();for (int i = 0;i<size;i++){if (i==size-1){res.offer(queue.peek().val);}TreeNode node = queue.poll();if (node.left!=null){queue.offer(node.left);}if (node.right!=null){queue.offer(node.right);}}}return res;} }

解題延申——求左視圖

求左視圖也非常簡單

if (i==size-1)? 這一行 改為?if (i==0)

即可。

解題本地調試代碼

??

import java.util.LinkedList; import java.util.List; import java.util.Queue;public class Solution199 {public static void main(String[] args) {String line = "[1,2,3,null,5,null,4]";TreeNode root = stringToTreeNode(line);List<Integer> retRight = new Solution199().rightSideView(root);List<Integer> retLeft = new Solution199().leftSideView(root);String outRight = integerArrayListToString(retRight);String outLeft = integerArrayListToString(retLeft);System.out.println("二叉樹為：" + line);System.out.println("右視圖為：" + outRight);System.out.println("左視圖為：" + outLeft);}public static TreeNode stringToTreeNode(String input) {input = input.trim();input = input.substring(1, input.length() - 1);if (input.length() == 0) {return null;}String[] parts = input.split(",");String item = parts[0];TreeNode root = new TreeNode(Integer.parseInt(item));Queue<TreeNode> nodeQueue = new LinkedList<>();nodeQueue.add(root);int index = 1;while (!nodeQueue.isEmpty()) {TreeNode node = nodeQueue.remove();if (index == parts.length) {break;}item = parts[index++];item = item.trim();if (!item.equals("null")) {int leftNumber = Integer.parseInt(item);node.left = new TreeNode(leftNumber);nodeQueue.add(node.left);}if (index == parts.length) {break;}item = parts[index++];item = item.trim();if (!item.equals("null")) {int rightNumber = Integer.parseInt(item);node.right = new TreeNode(rightNumber);nodeQueue.add(node.right);}}return root;}public static String integerArrayListToString(List<Integer> nums, int length) {if (length == 0) {return "[]";}StringBuilder result = new StringBuilder();for (int index = 0; index < length; index++) {Integer number = nums.get(index);result.append(Integer.toString(number)).append(", ");}return "[" + result.substring(0, result.length() - 2) + "]";}public static String integerArrayListToString(List<Integer> nums) {return integerArrayListToString(nums, nums.size());}public List<Integer> rightSideView(TreeNode root) {LinkedList<Integer> res = new LinkedList<>();if (root == null) {return res;}LinkedList<TreeNode> queue = new LinkedList<>();queue.offer(root);while (!queue.isEmpty()) {int size = queue.size();for (int i = 0; i < size; i++) {if (i == size - 1) {res.offer(queue.peek().val);}TreeNode node = queue.poll();if (node.left != null) {queue.offer(node.left);}if (node.right != null) {queue.offer(node.right);}}}return res;}public List<Integer> leftSideView(TreeNode root) {LinkedList<Integer> res = new LinkedList<>();if (root == null) {return res;}LinkedList<TreeNode> queue = new LinkedList<>();queue.offer(root);while (!queue.isEmpty()) {int size = queue.size();for (int i = 0; i < size; i++) {if (i == 0) {res.offer(queue.peek().val);}TreeNode node = queue.poll();if (node.left != null) {queue.offer(node.left);}if (node.right != null) {queue.offer(node.right);}}}return res;} }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;} }

輸出為

總結

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