Leetcode 617. 合并二叉树
题目:
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
题解:
public static TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if (t1 == null) {
return t2;
}
if (t2 == null) {
return t1;
}
TreeNode treeNode = new TreeNode(t1.value + t2.value);
Queue<TreeNode> queue = new LinkedList<>();
Queue<TreeNode> queue1 = new LinkedList<>();
Queue<TreeNode> queue2 = new LinkedList<>();
queue.add(treeNode);
queue1.add(t1);
queue2.add(t2);
while (!queue1.isEmpty() && !queue2.isEmpty()) {
TreeNode node = queue.poll();
TreeNode treeNode1 = queue1.poll();
TreeNode treeNode2 = queue2.poll();
TreeNode left1 = treeNode1.left;
TreeNode right1 = treeNode1.right;
TreeNode left2 = treeNode2.left;
TreeNode right2 = treeNode2.right;
if (left1 != null || left2 != null) {
if (left1 != null && left2 != null) {
TreeNode left = new TreeNode(left1.value + left2.value);
node.left = left;
queue.add(left);
queue1.add(left1);
queue2.add(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.value + right2.value);
node.right = right;
queue.add(right);
queue1.add(right1);
queue2.add(right2);
} else if (right1 != null) {
node.right = right1;
} else if (right2 != null) {
node.right = right2;
}
}
}
return treeNode;
}
public static TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if (t1 == null && t2 == null) return null;
if (t1 == null) return t2;
if (t2 == null) return t1;
Queue<TreeNode> q1 = new LinkedList<>();
Queue<TreeNode> q2 = new LinkedList<>();
q1.add(t1);
q2.add(t2);
while (!q1.isEmpty()) {
TreeNode n1 = q1.poll();
TreeNode n2 = q2.poll();
if (n1.left != null && n2.left != null) {
q1.add(n1.left);
q2.add(n2.left);
} else if (n1.left == null) {
n1.left = n2.left;
}
if (n1.right != null && n2.right != null) {
q1.add(n1.right);
q2.add(n2.right);
} else if (n1.right == null) {
n1.right = n2.right;
}
n1.value = n1.value + n2.value;
}
return t1;
}