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Monday, May 1, 2017

LeetCode OJ - Binary Tree Tilt


Please find the problem here.


It is obvious there is no way we can compute the tilt without walking the whole tree. Is it possible to do that in one pass? The answer is yes.


The key is to do the work in the stack unwind phase. Just do a normal recursive tree processing and return both the tilt and the sum of the subtree. That allows the unwind phase to compute the same two thing and return that back to the caller.

I hate passing by reference - that's why I insisted on using pointers - that notation make it blindingly obvious to the caller that the value will be changed. It would be much more implicit in the reference case.


#include "stdafx.h"


#include <map>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

    struct TreeNode
        int val;
        TreeNode *left;
        TreeNode *right;
        TreeNode(int x) : val(x), left(NULL), right(NULL) {}
    class Solution {
        void findTiltSum(TreeNode* root, int* pTilt, int* pSum)
            if (root == nullptr)
                *pTilt = 0;
                *pSum = 0;
                int leftTilt;
                int leftSum;
                int rightTilt;
                int rightSum;
                findTiltSum(root->left, &leftTilt, &leftSum);
                findTiltSum(root->right, &rightTilt, &rightSum);
                *pSum = leftSum + rightSum + root->val;
                *pTilt = abs(leftSum - rightSum) + leftTilt + rightTilt;

        int findTilt(TreeNode* root)
            int tilt;
            int sum;
            findTiltSum(root, &tilt, &sum);
            return tilt;

using namespace _LEET_BINARY_TREE_TILT;

    Solution solution;
    TreeNode a(1);
    TreeNode b(2);
    TreeNode c(3);
    a.left = &b;
    a.right = &c;
    b.left = nullptr;
    b.right = nullptr;
    c.left = nullptr;
    c.right = nullptr;
    cout << solution.findTilt(&a) << endl;
    return 0;

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