[刷题防痴呆] 0617 - 合并二叉树 (Merge Two

题目地址

https://leetcode.com/problems/merge-two-binary-trees/

题目描述

617. Merge Two Binary Trees

You are given two binary trees root1 and root2.

Imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge the two trees into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of the new tree.

Return the merged tree.

Note: The merging process must start from the root nodes of both trees.

 

Example 1:


Input: root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
Output: [3,4,5,5,4,null,7]
Example 2:

Input: root1 = [1], root2 = [1,2]
Output: [2,2]

思路

• dfs.

关键点

代码

• 语言支持：Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        TreeNode root = dfs(root1, root2);
        return root;
    }

    private TreeNode dfs(TreeNode node1, TreeNode node2) {
        if (node1 == null && node2 == null) {
            return null;
        }
        if (node1 == null) {
            return node2;
        }
        if (node2 == null) {
            return node1;
        }
        TreeNode node = new TreeNode();
        node.val = node1.val + node2.val;
        node.left = dfs(node1.left, node2.left);
        node.right = dfs(node1.right, node2.right);
        return node;
    }
}

class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) {
            return null;
        }
        if (root1 == null) {
            return root2;
        }
        if (root2 == null) {
            return root1;
        }
        int val = root1.val + root2.val;
        TreeNode node = new TreeNode(val);
        node.left = mergeTrees(root1.left, root2.left);
        node.right = mergeTrees(root1.right, root2.right);

        return node;
    }
}