LeetCode Add to List 617 Merge Two Binary Trees (深度优先搜索)

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

Example 1:

Input: Tree 1                     Tree 2                            1                         2                                      / \                       / \                                    3   2                     1   3                               /                           \   \                            5                             4   7                  Output: Merged tree:     3    / \   4   5  / \   \  5   4   7

Note: The merging process must start from the root nodes of both trees.

题目链接：https://leetcode.com/problems/merge-two-binary-trees/#/description

题目分析：DFS合并即可

击败77%的丑陋代码：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {        public void DFS(TreeNode t1, TreeNode t2, TreeNode ans) {        if (t1 != null && t2 != null) {            ans.val = t1.val + t2.val;            if (t1.left != null || t2.left != null) {                ans.left = new TreeNode(0);                DFS(t1.left, t2.left, ans.left);            }            if (t1.right != null || t2.right != null) {                ans.right = new TreeNode(0);                DFS(t1.right, t2.right, ans.right);            }            return;        } else if (t1 != null) {            ans.val = t1.val;            if (t1.left != null) {                ans.left = new TreeNode(0);                DFS(t1.left, null, ans.left);            }            if (t1.right != null) {                ans.right = new TreeNode(0);                DFS(t1.right, null, ans.right);            }        } else if (t2 != null) {            ans.val = t2.val;            if (t2.left != null) {                ans.left = new TreeNode(0);                DFS(null, t2.left, ans.left);            }            if (t2.right != null) {                ans.right = new TreeNode(0);                DFS(null, t2.right, ans.right);            }        }    }        public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {        TreeNode ans = null;        if (t1 == null && t2 == null) {            return ans;        }        ans = new TreeNode(0);        DFS(t1, t2, ans);        return ans;    }}

这样写更漂亮

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {        if (t1 == null && t2 == null) {            return null;        }        int val = (t1 == null ? 0 : t1.val) + (t2 == null ? 0 : t2.val);        TreeNode cur = new TreeNode(val);        cur.left = mergeTrees(t1 == null ? null : t1.left, t2 == null ? null : t2.left);        cur.right = mergeTrees(t1 == null ? null : t1.right, t2 == null ? null : t2.right);        return cur;    }}