LeetCode-437. Path Sum IIILevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/maximum-depth-of-binary-tree/

Given the root of a binary tree and an integer targetSum, return the number of paths where the sum of the values along the path equals targetSum.

The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).

Example 1:

Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
Output: 3
Explanation: The paths that sum to 8 are shown.

Example 2:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: 3

Constraints:

• The number of nodes in the tree is in the range [0, 1000].
• -10^9 <= Node.val <= 10^9
• -1000 <= targetSum <= 1000

【C++】

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

1. 双递归

小心数据溢出

class Solution {
public:
    int pathSum(TreeNode* root, int targetSum) {
        if (root == nullptr) {return 0;}
        return pathSumStartWithRoot(root, targetSum)
            + pathSum(root->left, targetSum)
            + pathSum(root->right, targetSum);
    }
 
    int pathSumStartWithRoot(TreeNode* root, int sum) {
        if (!root) {return 0;}
        int count = root->val == sum? 1: 0;
        if (double(sum) - double(root->val) < INT_MIN) {return 0;}
        int remaingSum = sum - root->val;
        count += pathSumStartWithRoot(root->left, remaingSum);
        count += pathSumStartWithRoot(root->right, remaingSum);
        return count;
    }
};

2.递归+回溯

class Solution {
public:
    int pathSum(TreeNode* root, int targetSum) {
        int pathCount = 0;
        vector<int> currentPath;
        pathSumHelper(root, targetSum, currentPath, pathCount);
        return pathCount;
    }
 
    void pathSumHelper(TreeNode *root,
                       int sum ,
                       vector<int> ¤tPath,
                       int &pathCount) {
        if(!root) return;
        currentPath.push_back(root->val);
        double currentSum = 0;
        for(int j=currentPath.size()-1; j>=0; j--) {
            currentSum += (double)currentPath[j];
            if(currentSum == sum) pathCount++;
        }
        pathSumHelper(root->left, sum, currentPath, pathCount);
        pathSumHelper(root->right, sum, currentPath, pathCount);
        currentPath.pop_back();
    }    
};

【Java】

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */

1. 双循环

class Solution {
    public int pathSum(TreeNode root, int targetSum) {
        if (root == null) {return 0;}
        return pathSumHelper(root, targetSum)
            + pathSum(root.left, targetSum)
            + pathSum(root.right, targetSum);
    }
    
    public int pathSumHelper(TreeNode root, int sum) {
        if (root == null) {return 0;}
        int count = sum == root.val ? 1 : 0;
        if (Double.valueOf(sum)
            - Double.valueOf(root.val)
            < Integer.MIN_VALUE) {
            return 0;
        }
        int remaining = sum - root.val;
        return count
            + pathSumHelper(root.left, remaining)
            + pathSumHelper(root.right, remaining);
    }
}

 参考文献

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