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解法

/**
 * 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) {
        // 这里判空很巧妙，一个为空就返回另一个，不用关注另一个是否为空
        // 能到后面，说明两个节点都不为空
        if (root1 == null) {
            return root2;
        }
        if (root2 == null) {
            return root1;
        }
        TreeNode root = new TreeNode(root1.val + root2.val);
        root.left = mergeTrees(root1.left, root2.left);
        root.right = mergeTrees(root1.right, root2.right);
        return root;        
    }
}