mergeTrees

Leetcode

题意：给定两课二叉树，然后融合成一颗。要求相同位置结点的值相加，没有子树的分支合并。
解题思路：以第一棵树为输出目标树，用递归的方法：若左子树为空而第二棵树的左子树不空，便接上第二棵树的左子树，右子树同样如此。若都有，则递归进入。递归出口是叶子结点。

/*
*File:mergeTrees.c:
*Author:0HP
*Date:20200222
*Purpose:to solve the problem in Leetcode
*https://leetcode.com/problems/merge-two-binary-trees/
*/
#include<stdio.h>
#include<stdlib.h>
struct TreeNode {
  int val;
  struct TreeNode *left;
  struct TreeNode *right;
};
void merge(struct TreeNode* t1,struct TreeNode* t2)
{
  //当前节点求和
  t1->val=t1->val+t2->val;
  //叶子结点，求和返回
  if(t2->left==NULL&&t2->right==NULL)
  {return;}
  if(t1->left!=NULL&&t2->left!=NULL)
  {merge(&(*t1->left),&(*t2->left));}
  if(t1->right!=NULL&&t2->right!=NULL)
  {merge(&(*t1->right),&(*t2->right));}
  //t1无左子树，t2有左子树
  if(t1->left==NULL&&t2->left!=NULL)
  {
    t1->left=t2->left;
  }
  //t1无右子树，t2有右子树
  if(t1->right==NULL&&t2->right!=NULL)
  {
    t1->right=t2->right;
  }
}
struct TreeNode* mergeTrees(struct TreeNode* t1, struct TreeNode* t2){
  if(t1==NULL){return t2;}
  if(t2==NULL){return t1;}
  merge(&(*t1),&(*t2));
  return t1;
}