617. Merge Two Binary Trees

For Tree 3, its each node's val is the sum(TreeNode t1 + TreeNode t2) in the same location.
Way to go deeper
if (t1 !=NULL && t2 != NULL) {
t1 = t1 + t2;
}

else if(t1 == NULL && t2 == NULL){return}

else{if(t1==NULL){t1 = t2;}else if(t2==NULL){t1 = t1;}}

mergeTrees(node.left);

mergeTrees(node.right);

每次return不知道return到哪儿去了

    if (t1 == null)
        return t2;
    if (t2 == null)
        return t1;

// check

    t1.val += t2.val;

// 赋值

    t1.left = mergeTrees(t1.left, t2.left);
    t1.right = mergeTrees(t1.right, t2.right);

// 继续调用，往下（recursive）

// 返回t1，给上一次调用他的mergeTrees((t1,t2).parent.left, (t1,t2).parent.left)
或者mergeTrees((t1,t2).parent.right, (t1,t2).parent.right)

SOLUTION

Approach #1 Using Recursion [Accepted]

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

Approach #2 Iterative Method [Accepted]