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 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]

.

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);
}