BiruLyu
5/30/2017 - 10:00 PM

102. Binary Tree Level Order Traversal(BFS).java

/**
 * 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<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> result = new ArrayList<>();
    levelOrderDFS(root, result, 0);
    return result;
  }

  private void levelOrderDFS(TreeNode root, List<List<Integer>> result, int level) {
    if (root == null) {
      return;
    }
    if (result.size() < level + 1) {
      List<Integer> list = new ArrayList<>();
      result.add(list);
    }
    result.get(level).add(root.val);
    levelOrderDFS(root.left, result, level + 1);
    levelOrderDFS(root.right, result, level + 1);
  }
}
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
 /*
 1. BFS 每次explored一层,for循环将一层的节点的子节点都放入queue,每次for循环结束level++
 2. BFS 将(TreeNode, Level)存入queue
 3. DFS 递归
        func(node, height) 
            result[height].add(node)
            func(node.left, heigh + 1)
            func(node.right, heigh + 1)
             
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (root == null) return res;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while(!queue.isEmpty()) {
            int curLen = queue.size();
            List<Integer> level = new ArrayList<Integer>();
            for(int i = 0; i < curLen; i++) {
                TreeNode temp = queue.poll();
                level.add(temp.val);
                if(temp.left != null) queue.offer(temp.left);
                if(temp.right != null) queue.offer(temp.right);
            }
            res.add(level);
        }
        return res;
    }
}