617. Merge Two Binary Trees（Easy）

标签：others   not   nod   while   递归   str   one   when   discuss   

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.

/*  合并的树是新建结点
    递归的思想，自己写不出来。。看discuss后觉得很简单，真尴尬,其实就是个二叉树的前序遍历
*/

/**
 * 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 && t2 == NULL)  return NULL;   
        TreeNode* root = new TreeNode((t1 != NULL ? t1 -> val: 0) +  (t2 != NULL ? t2 -> val: 0) );   // 初始化根节点，根据t1和t2的存在情况
        root -> left = mergeTrees((t1 != NULL ? t1 -> left : NULL), (t2 != NULL ? t2 -> left : NULL));
        root -> right = mergeTrees((t1 != NULL ? t1 -> right : NULL), (t2 != NULL ? t2 -> right : NULL));
        
        return root;
    }
};

617. Merge Two Binary Trees（Easy）

标签：others   not   nod   while   递归   str   one   when   discuss   

原文地址：http://www.cnblogs.com/simplepaul/p/7748103.html