【Leetcode】每日一題2021/10/08

• 思路

  我們讓一棵樹不動，讓另一棵樹加到上面去。并且要求兩者需要同步周遊相同的位置。為了保持順序，統一采用前序周遊。

  寫代碼的時候又是直接新定義一個樹，但是這樣寫出來判斷比較多，導緻耗時會更長。

// 建立一棵新樹
/**
 * 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;
        }
        // 中
        if (root1 != null && root2 == null) {
            TreeNode node = new TreeNode(root1.val);
            // 左
            node.left = mergeTrees(root1.left, null);
            // 右
            node.right = mergeTrees(root1.right, null);
            return node;
        }
        if (root1 == null && root2 != null) {
            TreeNode node = new TreeNode(root2.val);
            // 左
            node.left = mergeTrees(null, root2.left);
            // 右
            node.right = mergeTrees(null, root2.right);
            return node;
        }
        // if (root1 != null && root2 != null) {
        //     TreeNode node = new TreeNode(root1.val + root2.val);
        //     // 左
        //     node.left = mergeTrees(root1.left, root2.left);
        //     // 右
        //     node.right = mergeTrees(root1.right, root2.right);
        // }
        TreeNode node = new TreeNode(root1.val + root2.val);
        // 左
        node.left = mergeTrees(root1.left, root2.left);
        // 右
        node.right = mergeTrees(root1.right, root2.right);

        return node;
    }
}

// 參考代碼，精簡版
class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if (root1 == null) {
            return root2;
        }
        if (root2 == null) {
            return root1;
        }
        TreeNode node = new TreeNode(0);
        node.val = root1.val + root2.val;
        node.left = mergeTrees(root1.left, root2.left);
        node.right = mergeTrees(root1.right, root2.right);
        return node;
    }
}