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102. 二叉树的层序遍历

English Version

题目描述

给你一个二叉树，请你返回其按 层序遍历 得到的节点值。 （即逐层地，从左到右访问所有节点）。

 

示例：
二叉树：[3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回其层序遍历结果：

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

解法

队列实现。

Python3

# 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 root is None:
            return []
        res = []
        q = []
        q.append(root)
        while q:
            size = len(q)
            t = []
            for _ in range(size):
                node = q.pop(0)
                if node.left is not None:
                    q.append(node.left)
                if node.right is not None:
                    q.append(node.right)
                t.append(node.val)
            res.append(t)
        return res

Java

/**
 * 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 Collections.emptyList();
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        List<List<Integer>> res = new ArrayList<>();
        while (!q.isEmpty()) {
            int size = q.size();
            List<Integer> t = new ArrayList<>();
            while (size-- > 0) {
                TreeNode node = q.poll();
                t.add(node.val);
                if (node.left != null) q.offer(node.left);
                if (node.right != null) q.offer(node.right);
            }
            res.add(t);
        }
        return res;
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        if (!root) return {};
        vector<vector<int>> res;
        queue<TreeNode*> q{{root}};
        while (!q.empty()) {
            vector<int> oneLevel;
            for (int i = q.size(); i > 0; --i) {
                TreeNode* t = q.front();
                q.pop();
                oneLevel.push_back(t->val);
                if (t->left) q.push(t->left);
                if (t->right) q.push(t->right);
            }
            res.push_back(oneLevel);
        }
        return res;
    }
};

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