算法自由之路

102.二叉树的层序遍历

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

1、BFS 解决

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrder = function (root) {
  if (root == null) {
    return [];
  }
  const queue = [];
  const res = [];
  queue.push(root);
  while (queue.length) {
    let length = queue.length;
    const subList = [];
    for (let i = 0; i < length; i++) {
      const node = queue.shift();
      subList.push(node.val);
      if (node.left != null) {
        queue.push(node.left);
      }
      if (node.right != null) {
        queue.push(node.right);
      }
    }
    res.push(subList);
  }
  return res;
};

2、DFS

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrder = function (root) {
  const list = [];
  const helper = (list, root, level) => {
    if (root == null) {
      return;
    }
    if (level >= list.length) {
      list.push([]);
    }
    list[level].push(root.val);
    helper(list, root.left, level + 1);
    helper(list, root.right, level + 1);
  };
  helper(list, root, 0);
  return list;
};