数据结构定义：
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。
示例 1:
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
注意: 合并必须从两个树的根节点开始。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
递归方式：
class Solution {
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if(t1 == null && t2 == null)
return null;
TreeNode node = new TreeNode();
if(t1 != null)
node.val = t1.val;
if(t2 != null)
node.val += t2.val;
node.left = mergeTrees(t1 == null ? null: t1.left,
t2 == null ? null: t2.left);
node.right = mergeTrees(t1 == null ? null: t1.right,
t2 == null ? null: t2.right);
return node;
}
}
另一种递归改进方式：
/*
* 不是都创建一个新的节点
*/
class Solution {
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if(t1 == null)
return t2;
if(t2 == null)
return t1;
TreeNode node = new TreeNode(t1.val + t2.val);
node.left = mergeTrees(t1.left,t2.left);
node.right = mergeTrees(t1.right,t2.right);
return node;
}
}
广度优先遍历方式：
/*
* 思路： 定义三个队列，分别存储合并后的二叉树的节点以及两个原始二叉树的节点
* 定义一个节点，记录左右子树的连接情况
*/
class Solution {
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if(t1 == null)
return t2;
if(t2 == null)
return t1;
Queue<TreeNode> queue =new LinkedList<>();
Queue<TreeNode> queueT1 =new LinkedList<>();
Queue<TreeNode> queueT2 = new LinkedList<>();
TreeNode node = new TreeNode(t1.val + t2.val);
queue.offer(node);
queueT1.offer(t1);
queueT2.offer(t2);
while(!queueT2.isEmpty() && ! queueT1.isEmpty()){
TreeNode temp = queue.poll();
TreeNode node1 = queueT1.poll();
TreeNode node2 = queueT2.poll();
TreeNode left1 = node1.left,left2 = node2.left;
TreeNode right1 = node1.right,right2 = node2.right;
temp.left = getSubTreeNode(left1,left2,queue,queueT1,queueT2);
temp.right= getSubTreeNode(right1,right2,queue,queueT1,queueT2);
}
return node;
}
private TreeNode getSubTreeNode(TreeNode t1,TreeNode t2,
Queue<TreeNode> queue,
Queue<TreeNode> queueT1,
Queue<TreeNode> queueT2){
if(t1 != null && t2 != null){
TreeNode node = new TreeNode(t1.val + t2.val);
queueT1.offer(t1);
queueT2.offer(t2);
queue.offer(node);
return node;
}else{
return t1 != null ? t1 : t2;
}
}
}