1305. 两棵二叉搜索树中的所有元素

leetcode 链接

解法一：深度优先遍历二叉树

1、深度优先遍历二叉树

2、如果值为整数，则放入数组中

3、对数组排序

/**
 * 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} root1
 * @param {TreeNode} root2
 * @return {number[]}
 */
var getAllElements = function (root1, root2) {
  const list = [];
  function dfs(root) {
    if (root) {
      list.push(root.val);
      dfs(root.left);
      dfs(root.right);
    }
  }
  dfs(root1);
  dfs(root2);
  return list.sort((a, b) => a - b);
};

解法二：中序遍历 + 归并排序

先对二叉树中序遍历，得到两个有序数组，然后依次从两个数组头部取较小的数，最后得到一个升序数组。

/**
 * 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} root1
 * @param {TreeNode} root2
 * @return {number[]}
 */
var getAllElements = function (root1, root2) {
  const res = [];
  const list1 = [];
  const list2 = [];
  inorder(root1, list1);
  inorder(root2, list2);
  let i = 0,
    j = 0;
  while (i < list1.length && j < list2.length) {
    if (list1[i] < list2[j]) {
      res.push(list1[i]);
      i++;
    } else {
      res.push(list2[j]);
      j++;
    }
  }
  if (i === list1.length) {
    res.push(...list2.splice(j));
  }
  if (j === list2.length) {
    res.push(...list1.splice(i));
  }
  return res;
};

function inorder(root, list) {
  if (root) {
    inorder(root.left, list);
    list.push(root.val);
    inorder(root.right, list);
  }
}