102.binary-tree-level-order-traversal


102. 二叉树的层次遍历

给定一个二叉树,返回其按层次遍历的节点值。 (即逐层地,从左到右访问所有节点)。

例如:

给定二叉树: [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回其层次遍历结果:

[
  [3],
  [9,20],
  [15,7]
]

解法一

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
      if not root:
        return []

      result = []

      level = 0
      queue = []
      queue.append(root)
      while queue:
        result.append([])
        for i in range(len(queue)):
          node = queue.pop(0)
          result[level].append(node.val)

          if node.left:
            queue.append(node.left)
          if node.right:
            queue.append(node.right)

        level += 1
      return result
/**
 * 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) {
        if (root == null) return new ArrayList<>();

        int level = 0;
        List<List<Integer>> result = new ArrayList<>();
        LinkedList<TreeNode> queue = new LinkedList<>();

        queue.addFirst(root);
        while(!queue.isEmpty()) {
            result.add(new ArrayList<>());
            int size = queue.size();
            for(int i = 0; i < size; i++) {

                TreeNode node = queue.pop();
                result.get(level).add(node.val);

                if (node.left != null) queue.addLast(node.left);
                if (node.right != null) queue.addLast(node.right);
            }
            level++;
        }
        return result;
    }
}
/**
 * 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) {
        if (root == null) return new ArrayList<>();

        List<List<Integer>> result = new ArrayList<>();
        LinkedList<TreeNode> queue = new LinkedList<>();

        queue.addFirst(root);
        while(!queue.isEmpty()) {
            ArrayList<Integer> current = new ArrayList<>();
            int size = queue.size();
            for(int i = 0; i < size; i++) {
                TreeNode node = queue.pop();
                current.add(node.val);

                if (node.left != null) queue.addLast(node.left);
                if (node.right != null) queue.addLast(node.right);
            }
            result.add(current);
        }
        return result;
    }
}
/**
 * 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>> result = new ArrayList<>();
        if (root == null) return result;

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {

            List<Integer> list = new ArrayList<>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                list.add(node.val);
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
            result.add(list);
        }
        return result;
    }
}