一、leetcode617. 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]
由题意所得，有两种思路，dfs和bfs，dfs主靠递归，bfs主靠队列
dfs：
class Solution { public TreeNode mergeTrees(TreeNode root1, TreeNode root2) { if(root1 == null){return root2;} if(root2 == null){return root1;} TreeNode root = new TreeNode(root1.val+root2.val); root.left=mergeTrees(root1.left,root2.left); root.right=mergeTrees(root1.right,root2.right); return root; } }
bfs：
class Solution { public TreeNode mergeTrees(TreeNode t1, TreeNode t2) { if (t1 == null) { return t2; } if (t2 == null) { return t1; } TreeNode merged = new TreeNode(t1.val + t2.val); Queue<TreeNode> queue = new LinkedList<TreeNode>(); Queue<TreeNode> queue1 = new LinkedList<TreeNode>(); Queue<TreeNode> queue2 = new LinkedList<TreeNode>(); queue.offer(merged); queue1.offer(t1); queue2.offer(t2); while (!queue1.isEmpty() && !queue2.isEmpty()) { TreeNode node = queue.poll(), node1 = queue1.poll(), node2 = queue2.poll(); TreeNode left1 = node1.left, left2 = node2.left, right1 = node1.right, right2 = node2.right; if (left1 != null || left2 != null) { if (left1 != null && left2 != null) { TreeNode left = new TreeNode(left1.val + left2.val); node.left = left; queue.offer(left); queue1.offer(left1); queue2.offer(left2); } else if (left1 != null) { node.left = left1; } else if (left2 != null) { node.left = left2; } } if (right1 != null || right2 != null) { if (right1 != null && right2 != null) { TreeNode right = new TreeNode(right1.val + right2.val); node.right = right; queue.offer(right); queue1.offer(right1); queue2.offer(right2); } else if (right1 != null) { node.right = right1; } else { node.right = right2; } } } return merged; } }
