76 lines
2.4 KiB
Java
76 lines
2.4 KiB
Java
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//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。
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//
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// 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
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//。
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//
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// 示例 1:
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//
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//
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//输入:
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// Tree 1 Tree 2
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// 1 2
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// / \ / \
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// 3 2 1 3
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// / \ \
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// 5 4 7
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//输出:
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//合并后的树:
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// 3
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// / \
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// 4 5
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// / \ \
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// 5 4 7
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//
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//
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// 注意: 合并必须从两个树的根节点开始。
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// Related Topics 树 深度优先搜索 广度优先搜索 二叉树
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// 👍 724 👎 0
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package leetcode.editor.cn;
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import com.code.leet.entiy.TreeNode;
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//617:合并二叉树
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public class MergeTwoBinaryTrees {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new MergeTwoBinaryTrees().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
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TreeNode head;
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if (root1 != null && root2 != null) {
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head = new TreeNode(root1.val + root2.val);
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head.left = mergeTrees(root1.left, root2.left);
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head.right = mergeTrees(root1.right, root2.right);
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} else if (root1 == null && root2 == null) {
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head = null;
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} else if (root1 == null) {
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head = root2;
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} else {
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head = root1;
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}
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return head;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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