java数据结构与算法刷题-----LeetCode617. 合并二叉树

java数据结构与算法刷题目录（剑指Offer、LeetCode、ACM）-----主目录-----持续更新(进不去说明我没写完)：https://blog.youkuaiyun.com/grd_java/article/details/123063846

解题思路

1. 此题如果使用广度优先遍历，一定需要创建很多队列，代码量也会很多。所以选择深度优先。但是广度优先代码也会给出
2. 从根结点开始依次合并，如果当前合并位置有一个为null，那么直接等于另一个即可
3. 如果两个都不为null，那么将他俩的val值相加。最终返回相加后的结点

代码

1. 深度优先
   

/**
 * 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;//两个结点都为空，返回null
        if(root1 == null) return root2;//如果root1为空，那么直接返回root2，将root2放在当前应该合并位置
        if(root2 == null) return root1;//如果root2为空，那么直接返回root1，将root1放在当前应该合并位置
        //如果两个结点都存在
        root1.val += root2.val;//则合并
        root1.left = mergeTrees(root1.left,root2.left);//左子树合并
        root1.right = mergeTrees(root1.right,root2.right); //右子树合并
        return root1;//返回合并后的结点，最终递归完成后，一定返回根节点。
    }
}

2. 广度优先
   

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if (t1 == null) {
            return t2;
        }
        if (t2 == null) {
            return t1;
        }
        TreeNode merged = new TreeNode(t1.val + t2.val);
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        Queue<TreeNode> queue1 = new LinkedList<TreeNode>();
        Queue<TreeNode> queue2 = new LinkedList<TreeNode>();
        queue.offer(merged);
        queue1.offer(t1);
        queue2.offer(t2);
        while (!queue1.isEmpty() && !queue2.isEmpty()) {
            TreeNode node = queue.poll(), node1 = queue1.poll(), node2 = queue2.poll();
            TreeNode left1 = node1.left, left2 = node2.left, right1 = node1.right, right2 = node2.right;
            if (left1 != null || left2 != null) {
                if (left1 != null && left2 != null) {
                    TreeNode left = new TreeNode(left1.val + left2.val);
                    node.left = left;
                    queue.offer(left);
                    queue1.offer(left1);
                    queue2.offer(left2);
                } else if (left1 != null) {
                    node.left = left1;
                } else if (left2 != null) {
                    node.left = left2;
                }
            }
            if (right1 != null || right2 != null) {
                if (right1 != null && right2 != null) {
                    TreeNode right = new TreeNode(right1.val + right2.val);
                    node.right = right;
                    queue.offer(right);
                    queue1.offer(right1);
                    queue2.offer(right2);
                } else if (right1 != null) {
                    node.right = right1;
                } else {
                    node.right = right2;
                }
            }
        }
        return merged;
    }
}