Leetcode 617. Merge Two Binary Trees

题目链接：https://leetcode.cn/problems/merge-two-binary-trees/

参考链接：https://programmercarl.com/0617.%E5%90%88%E5%B9%B6%E4%BA%8C%E5%8F%89%E6%A0%91.html#%E5%85%B6%E4%BB%96%E8%AF%AD%E8%A8%80%E7%89%88%E6%9C%AC

方法一 递归-前序遍历

1 方法思想

2 代码实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
       public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) return null;
        TreeNode root = new TreeNode();
        if (root1 == null) {
            root.val = root2.val;
            root.left = root2.left;
            root.right = root2.right;
        }
        if (root2 == null) {
            root.val = root1.val;
            root.left = root1.left;
            root.right = root1.right;
        }
        if (root1 != null && root2 != null) {
            root.val = root1.val + root2.val;
            root.left = mergeTrees(root1.left, root2.left);
            root.right = mergeTrees(root1.right, root2.right);
        }
        return root;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37

3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

5 总结

方法一 栈

1 方法思想

2 代码实现

class Solution {
    // 使用栈迭代
    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;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36

3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

5 总结

方法三 队列

1 方法思想

2 代码实现

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;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35

3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

5 总结