Problem 199: Binary Tree Right Side View
思路
这道题其实就是在 BFS 遍历的基础上稍加改变:只存每个 level 最有边的那个数
怎么实现这个过程呢? 可以用 currIndex 标记当前的 node 的 index;用 nextIndex 标记下一个 node 的index
每次初始的时候,currIndex > 0, nextIndex = 0。这样做是因为每一个 level 上会有很多 node,当 currIndex 为 0 的时候,我们知道这是最后一个 node 了,可以存他了。
nextIndex 用来记下下一个 level 里有多少个 nodes,方便给下一次的 currIndex 赋值
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> rst = new ArrayList<Integer>();
if (root == null) {
return rst;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
int currIndex = 1, nextIndex = 0;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode currNode = queue.poll();
currIndex--;
if (currIndex == 0) {
rst.add(currNode.val);
}
if (currNode.left != null) {
queue.offer(currNode.left);
nextIndex++;
}
if (currNode.right != null) {
queue.offer(currNode.right);
nextIndex++;
}
if (currIndex == 0) {
currIndex = nextIndex;
nextIndex = 0;
}
}
}
return rst;
}
}
易错点
如果是求 left side view 呢?
if (nextIndex == 0) { rst.add(currNode.val); }
只需要做这一个改动。
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