617. Merge Two Binary Trees

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本文介绍了一道LeetCode上的题目——合并两棵二叉树的方法。当两棵树有重叠节点时，将节点值相加作为新节点的值；如果没有重叠，则使用非空节点作为新树的节点。通过递归方式实现合并过程。

题目来源【Leetcode】

Given two binary trees and 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 them 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 new tree.

Example 1:
Input:
Tree 1 Tree 2
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
Output:
Merged tree:
3
/ \
4 5
/ \ \
5 4 7
Note: The merging process must start from the root nodes of both trees.

这道题是合并两个二叉树，代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if(t1 == NULL ) return t2;
        if(t2 == NULL) return t1;
        TreeNode*root = new TreeNode((t1?t1->val:0)+(t2?t2->val:0));
        root->left = mergeTrees((t1?t1->left:NULL),(t2?t2->left:NULL));
        root->right = mergeTrees((t1?t1->right:NULL),(t2?t2->right:NULL));
        return root;
    }
};