力扣617题（合并二叉树）

617、合并二叉树

基本思想：

具体实现：

1、确定参数和返回值

参数：两个二叉树的根节点

返回值：合并之后新二叉树的根节点

2、确定终止条件

遍历到叶子节点

3、单层递归逻辑

TreeNode newRoot = new TreeNode(root1.val + root2.val);
newRoot.left = mergeTrees(root1.left, root2.left);
newRoot.right = mergeTrees(root1.right, root2.right);

代码：

class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if (root1 == null) return root2;
        if (root2 == null) return root1;

        TreeNode newRoot = new TreeNode(root1.val + root2.val);
        newRoot.left = mergeTrees(root1.left, root2.left);
        newRoot.right = mergeTrees(root1.right, root2.right);
        return newRoot;
    }
}

使用栈迭代

class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if (root1 == null) return root2;
        if (root2 == null) return root1;

        Stack<TreeNode> stack = new Stack<>();
        stack.push(root2);
        stack.push(root1);
        while (!stack.isEmpty()){
            TreeNode node1 = stack.pop();
            TreeNode node2 = stack.pop();
            node1.val += node2.val;
            if (node2.right != null && node1.right != null) {
                stack.push(node2.right);
                stack.push(node1.right);
            } else {
                if (node1.right == null) node1.right = node2.right;
            }

            if (node2.left != null && node1.left != null){
                stack.push(node2.left);
                stack.push(node1.left);
            } else {
                if (node1.left == null) {
                    node1.left = node2.left;
                }
            }
        }
        return root1;
    }
}

使用队列迭代

class Solution {
    // 使用队列迭代
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if (root1 == null) return root2;
        if (root2 ==null) return root1;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root1);
        queue.offer(root2);
        while (!queue.isEmpty()) {
            TreeNode node1 = queue.poll();
            TreeNode node2 = queue.poll();
            // 此时两个节点一定不为空，val相加
            node1.val = node1.val + node2.val;
            // 如果两棵树左节点都不为空，加入队列
            if (node1.left != null && node2.left != null) {
                queue.offer(node1.left);
                queue.offer(node2.left);
            }
            // 如果两棵树右节点都不为空，加入队列
            if (node1.right != null && node2.right != null) {
                queue.offer(node1.right);
                queue.offer(node2.right);
            }
            // 若node1的左节点为空，直接赋值
            if (node1.left == null && node2.left != null) {
                node1.left = node2.left;
            }
            // 若node2的左节点为空，直接赋值
            if (node1.right == null && node2.right != null) {
                node1.right = node2.right;
            }
        }
        return root1;
    }
}