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.
將給定的兩個(gè)二叉樹(shù)合并水慨，對(duì)應(yīng)節(jié)點(diǎn)都存在的話將值相加作為合并后的節(jié)點(diǎn)值。
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.
思路
【方法1】遞歸
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
if(!t1) return t2;
if(!t2) return t1;
t1->val += t2->val;
t1->left = mergeTrees(t1->left,t2->left);
t1->right = mergeTrees(t1->right,t2->right);
return t1;
}
};
/**
* 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) return t2;
if(t2==null) return t1;
t1.val += t2.val;
t1.left = mergeTrees(t1.left,t2.left);
t1.right = mergeTrees(t1.right,t2.right);
return t1;
}
}
