No.617 合并二叉树

给你两棵二叉树： root1 和 root2 。

想象一下，当你将其中一棵覆盖到另一棵之上时，两棵树上的一些节点将会重叠（而另一些不会）。你需要将这两棵树合并成一棵新二叉树。合并的规则是：如果两个节点重叠，那么将这两个节点的值相加作为合并后节点的新值；否则，不为 null 的节点将直接作为新二叉树的节点。

返回合并后的二叉树。

注意: 合并过程必须从两个树的根节点开始。

示例 1：

输入：root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
输出：[3,4,5,5,4,null,7]

示例 2：

输入：root1 = [1], root2 = [1,2]
输出：[2,2]

提示：

• 两棵树中的节点数目在范围 [0, 2000] 内
• -10^4 <= Node.val <= 10^4

解题思路

实现

深度优先搜索 - 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} root1
 * @param {TreeNode} root2
 * @return {TreeNode}
 */
var mergeTrees = function (root1, root2) {
  if (!root1) return root2;
  if (!root2) return root1;
  let merged = new TreeNode(root1.val + root2.val);
  merged.left = mergeTrees(root1.left, root2.left);
  merged.right = mergeTrees(root1.right, root2.right);
  return merged;
};

广度优先搜索 - bfs

/**
 * 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 {TreeNode}
 */
var mergeTrees = function (root1, root2) {
  if (!root1) return root2;
  if (!root2) return root1;
  let queue = [];
  let queue1 = [];
  let queue2 = [];
  let merged = new TreeNode(root1.val + root2.val);
  queue.push(merged);
  queue1.push(root1);
  queue2.push(root2);
  while (queue1.length && queue2.length) {
    let node = queue.shift();
    let node1 = queue1.shift();
    let node2 = queue2.shift();
    let left1 = node1.left;
    let left2 = node2.left;
    let right1 = node1.right;
    let right2 = node2.right;
    if (left1 || left2) {
      if (left1 && left2) {
        let left = new TreeNode(left1.val + left2.val);
        node.left = left;
        queue.push(left);
        queue1.push(left1);
        queue2.push(left2);
      } else if (left1) {
        node.left = left1;
      } else if (left2) {
        node.left = left2;
      }
    }
    if (right1 || right2) {
      if (right1 && right2) {
        let right = new TreeNode(right1.val + right2.val);
        node.right = right;
        queue.push(right);
        queue1.push(right1);
        queue2.push(right2);
      } else if (right1) {
        node.right = right1;
      } else if (right2) {
        node.right = right2;
      }
    }
  }
  return merged;
};