给定两个二叉树，想象当你将它们中的一个覆盖到另一个上时，两个二叉树的一些节点便会重叠。
你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠，那么将他们的值相加作为节点合并后的新值，否则不为 NULL 的节点将直接作为新二叉树的节点。
输入:
Tree 1 Tree 2
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
输出:
合并后的树:
3
/ \
4 5
/ \ \
5 4 7
let mergeTrees = (l1, l2) => {
if (l1 === null && l2 === null)
return null;
if (l1 === null)
return l2;
if (l2 === null)
return l1;
let root = TreeNode(l1.val + l2.val);
root.left = mergeTrees(l1.left, l2.left);
root.right = mergeTrees(l1.right, l2.right);
return root;
};func mergeTrees(t1 *TreeNode, t2 *TreeNode) *TreeNode {
if t1 == nil && t2 == nil {
return nil
}
if t1 == nil {
return t2
}
if t2 == nil {
return t1
}
root := &TreeNode{Val: t1.Val + t2.Val}
root.Left = mergeTrees(t1.Left, t2.Left)
root.Right = mergeTrees(t1.Right, t2.Right)
return root
}class Solution {
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if (t1 == null && t2 == null) return null;
if (t1 == null) return t2;
if (t2 == null) return t1;
TreeNode root = new TreeNode(t1.val + t2.val);
root.left = mergeTrees(t1.left, t2.left);
root.right = mergeTrees(t1.right, t2.right);
return root;
}
}