LeetCode-199. Binary Tree Right Side View (JAVA)（二叉树最右侧结点)

199. Binary Tree Right Side View

Given a binary tree, imagine yourself standing on theright 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].

bfs

返回二叉树的最右侧结点，BFS记录最后一个结点，

从右边看，并不是指右节点--也就是每层最右边的节点，则考虑层次遍历只取最右

LeetCode-515. Find Largest Value in Each Tree Row (JAVA)（二叉树每行的最大值）

public List<Integer> rightSideView(TreeNode root) {List<Integer> ret = new ArrayList<>();if (root == null)return ret;bfs(root, ret);return ret;}private void bfs(TreeNode root, List<Integer> ret) {Queue<TreeNode> q = new LinkedList<>();q.add(root);// 层次遍历，只需要记录本层结点个数即可int curNum = 1;// 下一层的节点数int nextNum = 0;while (!q.isEmpty()) {TreeNode node = q.poll();if (curNum == 1)ret.add(node.val);curNum--;if (node.left != null) {q.offer(node.left);nextNum++;}if (node.right != null) {q.offer(node.right);nextNum++;}if (curNum == 0) {curNum = nextNum;nextNum = 0;}}}

和上题目一样使用前序遍历递归解法

dfs(preorder)

public List<Integer> rightSideView(TreeNode root) {List<Integer> ret = new ArrayList<>();rightSideView(root, 0, ret);return ret;}public void rightSideView(TreeNode root, int level, List<Integer> ret) {if (root == null)return;// 当等于size的时候说明是最右侧结点，直接加入即可if (ret.size() == level)ret.add(root.val);rightSideView(root.right, level + 1, ret);rightSideView(root.left, level + 1, ret);}