LeetCode Top100 合并二叉树

 

 

递归解法：

// 不修改原二叉树做法
class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if(t1 == null || t2 == null)
            return t1 == null ? t2 : t1;
        // 不修改原二叉树做法
        TreeNode res = new TreeNode(t1.val + t2.val);
        
        res.left = mergeTrees(t1.left, t2.left);
        res.right = mergeTrees(t1.right, t2.right);

        return res;
    } 
}


//结果合并在t1上

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

        return t1;
    } 
}

 

非递归解法：用队列，一对一对的入队，如果左边为空，则把第二个树的子树直接赋给左边的树，下面就不管它

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if (t1 == null && t2 == null) return null;
        if (t1 == null || t2 == null) 
            return t1 == null ? t2 : t1;
        LinkedList queue = new LinkedList<>();
        queue.add(t1);
        queue.add(t2);
        while (!queue.isEmpty()){
            //出队
            TreeNode l1 = queue.remove();
            TreeNode l2 = queue.remove();
            l1.val += l2.val;
            if (l1.left == null){
                l1.left = l2.left;   
            }
            else if (l1.left != null && l2.left != null){
                queue.offer(l1.left);
                queue.offer(l2.left);  
            }
            
            if (l2.right == null){
                l1.right = l2.right;   
            }
            else if (l1.right != null && l2.right != null){
                queue.offer(l1.right);
                queue.offer(l2.right);  
            }
        }
        return t1;
    } 
}

更加简洁的写法，队列里存大小为2的数组

public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if (t1 == null)
            return t2;
        if (t2 == null)
            return t1;
        // use array in the queue to manipulate at the same time
        Queue queue = new LinkedList<>();
        queue.offer(new TreeNode[]{t1,t2});
        
        while (!queue.isEmpty()){
            TreeNode[] nodes = queue.poll();
            // merge 2 into 1 when it is not null
            if (nodes[1]  == null)
                continue;
            // nodes[0] must not be null
            nodes[0].val += nodes[1].val;
            // make sure nodes[0] will be null
            if (nodes[0].left == null)
                nodes[0].left = nodes[1].left;
            else
                queue.offer(new TreeNode[]{nodes[0].left,nodes[1].left});
            if (nodes[0].right == null)
                nodes[0].right = nodes[1].right;
            else
                queue.offer(new TreeNode[]{nodes[0].right,nodes[1].right});

        }
        return t1;