102. Binary Tree Level Order Traversal
这是一道经典 BFS 题目。
Tree related problem + address relationship on the same level -> 通常想到 BFS。用 Queue 实现。
expand(1)-> generate(3) and generate(2)
操作过程：1) Expand a node s 2) Generate s’s neighbor node: reach out to its nrighboring node: First, generate node 3, and then generate node 2.
数据结构：Queue (FIFO) (Termination condition is when the queue is empty.)
Time=O(n), Space=O(n)。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> cur = new ArrayList<>();
for (int i = 0; i < size; i ++) {
TreeNode curNode = queue.poll();
cur.add(curNode.val);
if (curNode.left != null) {
queue.offer(curNode.left);
}
if (curNode.right != null) {
queue.offer(curNode.right);
}
}
res.add(cur);
}
return res;
}
}
本题也可以用 DFS 来做。Time=O(n), Space=O(n)。
/**
* 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<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
helper(root, 0, res);
return res;
}
private void helper(TreeNode node, int level, List<List<Integer>> res) {
if (res.size() == level) {
res.add(new ArrayList<>());
}
res.get(level).add(node.val);
if (node.left != null) {
helper(node.left, level + 1, res);
}
if (node.right != null) {
helper(node.right, level + 1, res);
}
}
}