617. Merge Two Binary Trees

最新推荐文章于 2020-05-10 14:18:51 发布

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本文介绍了一种将两棵二叉树合并为一棵新二叉树的算法。合并时，若有节点重叠，则求和作为新节点值；若不重叠，则保留非空节点。通过递归方式实现前序遍历，确保从根节点开始合并。

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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.

解决思路：

从头到尾对树遍历，先序遍历。

/**
 * 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) {
        TreeNode* merge_tree = NULL;
        if (t1 && !t2){
            merge_tree = new TreeNode(t1->val);
            merge_tree->left = mergeTrees(t1->left, NULL);
            merge_tree->right = mergeTrees(t1->right, NULL);
        }    
        if (!t1 && t2){
            merge_tree = new TreeNode(t2->val);
            merge_tree->left = mergeTrees(NULL, t2->left);
            merge_tree->right = mergeTrees(NULL, t2->right);
        }  
        if(t1 && t2){
            merge_tree = new TreeNode(t1->val + t2->val);
            merge_tree->left = mergeTrees(t1->left, t2->left);
            merge_tree->right = mergeTrees(t1->right, t2->right);
        }
        return merge_tree;
    }
};

question：以上递归剪枝存在问题，应该及时return，减少判断次数。

/**
 * 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) {
        TreeNode* merge_tree = NULL;
        if (t1 && !t2){
            merge_tree = new TreeNode(t1->val);
            merge_tree->left = mergeTrees(t1->left, NULL);
            merge_tree->right = mergeTrees(t1->right, NULL);
            return merge_tree;
        }    
        if (!t1 && t2){
            merge_tree = new TreeNode(t2->val);
            merge_tree->left = mergeTrees(NULL, t2->left);
            merge_tree->right = mergeTrees(NULL, t2->right);
            return merge_tree;
        }  
        if(t1 && t2){
            merge_tree = new TreeNode(t1->val + t2->val);
            merge_tree->left = mergeTrees(t1->left, t2->left);
            merge_tree->right = mergeTrees(t1->right, t2->right);
            return merge_tree;
        }
        return merge_tree;
    }
};