LeetCode : 617. Merge Two Binary Trees 合并二叉树

Love The Way You Lie 2021-06-24 16:11 335阅读 0赞

**试题**  
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

**代码**

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