Easy_MergeBinaryTree_617.java

package Leetcode;

import apple.laf.JRSUIUtils.Tree;

/*
    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.
 */
public class Easy_MergeBinaryTree_617 {

    public static void main(String[] args){

    }

    // Runtime: 12 ms, faster than 20.05% of Java online submissions
    // 很普通的递归思路
    private static TreeNode mergeTrees(TreeNode t1, TreeNode t2){
        if(t1 == null && t2 == null) {return null;}
        if(t2 != null && 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;
    }
    // Runtime: 8ms, a simpler logic
    private static TreeNode mergeTrees_v2(TreeNode t1, TreeNode t2){
        if(t1 == null || t2 == null){
            return t1 == null ? t2 : t1;
        }else{
            t1.val += t2.val;
            t1.left = mergeTrees_v2(t1.left, t2.left);
            t1.right = mergeTrees_v2(t1.right, t2.right);
            return t1;
        }
    }
}

class TreeNode{

    public int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x){
        this.val = x;
    }

}