leetcode算法題2: 合併兩個二叉樹。遞歸，如何切入並保持清醒？

/*
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.
*/

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* mergeTrees(struct TreeNode* t1, struct TreeNode* t2) {
    
}

題意：合併兩個二叉樹。規則是，對應的兩個結點若是有重疊就求和（好比更改val），若是不重疊就使用非空的結點的值。（不考慮樹釋放的問題。）

辦法1

* 把t2併到t1，最終返回t1。

* 記錄t1，做爲最終的返回。進入遞歸函數。

* 若是t1與t2都不爲空，此時須要遞歸：

   * 求和，更改t1->val。

   * 左子樹遞歸，並把結果賦值給t1->left。

   * 右子樹遞歸，並把結果賦值給t1->right。

   * 返回t1。

* 不然，結束遞歸，返回t1?t1:t2。

* 注意，遞歸的返回值要利用上，這是更換子樹的機會。

強化與提問

* 遞歸關鍵點：

   * 只考慮兩層，root跟（root->left && root->right）。

   * 什麼時候遞歸。

   * 什麼時候結束遞歸。

* 方法不同時，要留意的點極可能也不同。在一個方法中容易犯的錯，在另外一個方法中可能根本就不存在。因此，想辦法有時候是解決問題的關鍵。

* 代碼上的細節，有時決定了成敗。

---

#include <stdio.h>

struct TreeNode {
  int val;
  struct TreeNode *left;
  struct TreeNode *right;
};

struct TreeNode* merge(struct TreeNode* l, struct TreeNode* r) {
    if (l && r) {
        l->val += r->val;
        l->left = merge(l->left, r->left);
        l->right = merge(l->right, r->right);
        return l;
    }
    else {
        return l?l:r;
    }
}
 
struct TreeNode* mergeTrees(struct TreeNode* t1, struct TreeNode* t2) {
   struct TreeNode* ret = t1;
   merge(t1, t2);
   return ret;
}

void printTree(struct TreeNode* tree) {
    if (tree) {
        printf("%d, ", tree->val);
        printTree(tree->left);
        printTree(tree->right);
    }
    else {
        printf("NULL, ");
    }
}

int main(int argc, char *argv[])
{
    struct TreeNode n1={1,0,0};
    struct TreeNode n2={3,0,0};
    struct TreeNode n3={2,0,0};
    struct TreeNode n4={5,0,0};
    n1.left = &n2;
    n1.right = &n3;
    n2.left = &n4;
    struct TreeNode r1={2,0,0};
    struct TreeNode r2={1,0,0};
    struct TreeNode r3={3,0,0};
    struct TreeNode r4={4,0,0};
    struct TreeNode r5={7,0,0};
    r1.left = &r2;
    r1.right = &r3;
    r2.right = &r4;
    r3.right = &r5;
    printf("tree1:\n");
    printTree(&n1);
    printf("\ntree2:\n");
    printTree(&r1);
    printf("\nmerged:\n");
    struct TreeNode* ret = mergeTrees(&n1, &r1);
    printTree(ret);
    printf("\n");

    return 0;
}