617. Merge Two Binary Trees（二叉树的合并）

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.

代码参考：http://blog.csdn.net/wchstrife/article/details/73290593

/** * 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) {            if(t1==null && t2==null){                                    return null;              }              if(t1==null && t2!=null){                                    return t2;              }              if(t1!=null && t2==null){                                    return t1;              }              if(t1!=null && t2!=null){                  t1.val += t2.val;  //              采用递归依次分析                t1.left = mergeTrees(t1.left, t2.left);                  t1.right = mergeTrees(t1.right, t2.right);              }                            return t1;          }      }

方法一：通过时间14ms

二叉树只有四种可能，通过返回二叉树t1来得到结果，t1的结果可以通过递归来得到。这种解法很巧妙。