563. Binary Tree Tilt

Given a binary tree, return the tilt of the whole tree.

The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.

The tilt of the whole tree is defined as the sum of all nodes' tilt.

Example:

Input: 1 / \ 2 3 Output: 1 Explanation: Tilt of node 2 : 0 Tilt of node 3 : 0 Tilt of node 1 : |2-3| = 1 Tilt of binary tree : 0 + 0 + 1 = 1

Note:

The sum of node values in any subtree won't exceed the range of 32-bit integer. All the tilt values won't exceed the range of 32-bit integer.

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思路：

代码1：

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int findTilt(TreeNode* root) { if(!root) {return 0;} if(!root->left && !root->right){return 0;} int leftval=0,rightval=0,dif=0; if(root->left){ dif += findTilt(root->left); leftval = root->left->val; } if(root->right){ dif += findTilt(root->right); rightval =root->right->val; } dif += abs(leftval-rightval); root->val += (leftval+rightval); return dif; } };

代码2：

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int res=0; int findTil(TreeNode* root) { if(!root) {return 0;} int leftsum,rightsum,sum; if(root->left!=NULL){leftsum=findTil(root->left);} else{leftsum=0;} if(root->right!=NULL){rightsum=findTil(root->right);} else{rightsum=0;} sum=leftsum+rightsum+root->val; res += abs(leftsum-rightsum); return sum; } int findTilt(TreeNode* root) { int temp=findTil(root); return res; } };

代码3：