【LeetCode】617. Merge Two Binary Trees

afra發表於2019-05-11
LeetCode

題目描述

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.

解決方案

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

解題思路
遍歷二叉樹的思想，從根節點開始，同時遞迴遍歷兩棵二叉樹，遍歷過程中合併兩棵二叉樹。注意空節點的判斷。

Submission Details

183 / 183 test cases passed.
Status: Accepted
Runtime: 49 ms

解決方案-改良

/**
 * 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* t;
        if(t1 && t2){
            t = new TreeNode(t1->val + t2->val);
            t->left = mergeTrees(t1->left, t2->left);
            t->right = mergeTrees(t1->right, t2->right);
            return t;
        }else {
            return t1 ? t1 : t2;
        }
    } 
};

Submission Details

183 / 183 test cases passed.
Status: Accepted
Runtime: 39 ms
beats 91.78 % of cpp submissions

617. Merge Two Binary Trees

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