Leetcode 617. Merge Two Binary Trees.（Java）

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本文介绍了一种将两棵二叉树合并为一棵新二叉树的算法，当两棵树的节点重叠时，新树的节点值为两节点值之和；若仅有一棵树的节点存在，则该节点成为新树的对应节点。采用递归方式实现，从根节点开始逐层处理。

问题描述

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.
Example1
Note: The merging process must start from the root nodes of both trees.

解决方案

package test ;

import java.lang.* ;
import java.util.*;

//  Definition for a binary tree node.
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; } //构造函数
}

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        TreeNode result ;
        if(t1 == null && t2 == null){
            return null ;
        }
        if(t1 == null && t2 != null){
            return t2 ;
        }
        if(t1 != null && t2 == null){
            return t1 ;
        }

        result = new TreeNode(t1.val + t2.val) ;
        result.left = mergeTrees(t1.left, t2.left) ;
        result.right = mergeTrees(t1.right, t2.right) ;

        return result ;
    }

    public void printTree(TreeNode t){
        if(t != null)
        {
            System.out.println(t.val) ;
            printTree(t.left) ;
            printTree(t.right) ;
        }
    }
}

public class Test{
    public static void main(String[] args){
        Solution s = new Solution() ;
        TreeNode te1 = new TreeNode(1) ;
        TreeNode te11 = new TreeNode(3) ;
        TreeNode te12 = new TreeNode(2) ;
        TreeNode te13 = new TreeNode(5) ;
        te1.left = te11 ; te1.right = te12 ; te11.left = te13 ;

        TreeNode te2 = new TreeNode(2) ;
        TreeNode te21 = new TreeNode(1) ;
        TreeNode te22 = new TreeNode(3) ;
        TreeNode te23 = new TreeNode(4) ;
        TreeNode te24 = new TreeNode(7) ;
        te2.left = te21 ; te2.right = te22 ; te21.right = te23 ; te22.right = te24 ;

        s.printTree(s.mergeTrees(te1, te2));
    }
}