617 合并二叉树(带有返回值)

给定两个二叉树，想象当你将它们中的一个覆盖到另一个上时，两个二叉树的一些节点便会重叠。

你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠，那么将他们的值相加作为节点合并后的新值，否则不为 NULL 的节点将直接作为新二叉树的节点。

示例 1:

输入: 
    Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
输出: 
合并后的树:
         3
        / \
       4   5
      / \   \ 
     5   4   7

注意: 合并必须从两个树的根节点开始。

我没绕过来的点：当两个数节点都有值的时候，需要返回两个节点值的和；递归只有当两者都有值的时候才能继续往下进行。如果一旦某个地方只有一个值，那么返回这个值（注意，此时这个值是树的节点，和链表一样，是包含它的两个子节点的，比如两个左下链表的树和一个右下链表的树）

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        TreeNode root = null;
        if(t1 == null && t2 == null)
            return root;
        if(t1 == null && t2 != null)
           return t2;                   
        if(t1 != null && t2 == null)
            return t1;            

        root = new TreeNode(t1.val + t2.val);
        root.left = mergeTrees(t1.left,t2.left);
        root.right = mergeTrees(t1.right,t2.right);       

        return root;
    }
}

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if(t1 == null && t2 == null)
            return null;

        if(t1 != null && t2 != null){
            t1.val += t2.val; 
            t1.left = mergeTrees(t1.left,t2.left);
            t1.right = mergeTrees(t1.right,t2.right);
        }
        if(t1 == null && t2 != null){
            t1 = new TreeNode(t2.val);
            t1.left = mergeTrees(null,t2.left);
            t1.right = mergeTrees(null,t2.right);
        }
        if(t1 != null && t2 == null){
            t1.left = mergeTrees(t1.left,null);
            t1.right = mergeTrees(t1.right,null);
        }

        return t1;
    }
}

犯的错误：没有添加t1.left = …和t1.right=…

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        TreeNode root = null;
        if(t1 == null && t2 == null)
            return root;
        if(t1 == null && t2 != null){
            root = new TreeNode(t2.val);
            root.left = mergeTrees(null,t2.left);
            root.right = mergeTrees(null,t2.right);
        }
        if(t1 != null && t2 == null){
             root = new TreeNode(t1.val);
             root.left = mergeTrees(t1.left,null);
             root.right = mergeTrees(t1.right,null);
        }
        if(t1 != null && t2 != null){
            root = new TreeNode(t1.val + t2.val);
            root.left = mergeTrees(t1.left,t2.left);
            root.right = mergeTrees(t1.right,t2.right);
        }

        return root;
    }
}

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        TreeNode root = null;
        if(t1 == null && t2 == null)
            return root;
        if(t1 == null && t2 != null){
            root = t2;           
        }
        if(t1 != null && t2 == null){
             root = t1;            
        }
        if(t1 != null && t2 != null){
            root = new TreeNode(t1.val + t2.val);
            root.left = mergeTrees(t1.left,t2.left);
            root.right = mergeTrees(t1.right,t2.right);
        }

        return root;
    }
}