[leetcode-663-Equal Tree Partition]

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Given a binary tree with n nodes, your task is to check if it‘s possible to partition the tree to two trees which have the equal sum of values after removing exactly one edge on the original tree.

Example 1:

Input:     
    5
   /   10 10
    /     2   3

Output: True
Explanation: 
    5
   / 
  10
      
Sum: 15

   10
  /   2    3

Sum: 15

Example 2:

Input:     
    1
   /   2  10
    /     2   20

Output: False
Explanation: You can‘t split the tree into two trees with equal sum after removing exactly one edge on the tree.

Note:

1. The range of tree node value is in the range of [-100000, 100000].
2. 1 <= n <= 10000

思路：

递归求子树的和，观察子树的和是否是整个树的和的一半即可。

int sumTree(TreeNode* root)
{
  if(root==NULL)return 0;
  int left = sumTree(root->left);  
  int right = sumTree(root->right);
  return root->val+left+right;
}

bool bianli(TreeNode* root,int sum)
{
  if(root==NULL)return false;
  if(sumTree(root->left)*2==sum)return true;  
  if(sumTree(root->right)*2==sum)return true;
  return (bianli(root->left,sum) ||  bianli(root->right,sum));
}
 bool checkEqualTree(TreeNode* root)
 {
   if(root==NULL ||(root->left==NULL && root->right==NULL))return false;
    int sum = sumTree(root);
    if(sum%2==1)return false;
     return bianli(root,sum);        
  }

[leetcode-663-Equal Tree Partition]

标签：ati   style   bsp   div   pre   note   art   etc   tree node   

原文地址：http://www.cnblogs.com/hellowooorld/p/7399292.html