文章作者：Tyan
博客：noahsnail.com  |  CSDN  |  简书

1. Description

Path Sum

2. Solution

2.1 Recursive

• Version 1.0

/**
 * 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:
    bool hasPathSum(TreeNode* root, int sum) {
        if(root == NULL) {
            return false;
        }
        return tranverseTree(root, 0, sum);
    }
    
    bool tranverseTree(TreeNode* root, int currentSum, int sum) {
        currentSum += root->val;
        if(currentSum == sum && root->left == NULL && root->right == NULL) {
            return true;
        }
        if(root->left == NULL && root->right == NULL) {
            return false;
        }
        if(root->left != NULL && root->right != NULL) {
            return tranverseTree(root->left, currentSum, sum) || tranverseTree(root->right, currentSum, sum);
        }
        if(root->left != NULL) {
            return tranverseTree(root->left, currentSum, sum);
        }
        else {
            return tranverseTree(root->right, currentSum, sum);
        }
    }
};

• Version 2.0

/**
 * 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:
    bool hasPathSum(TreeNode* root, int sum) {
        if(root == NULL) {
            return false;
        }
        if(root->val == sum && root->left == NULL && root->right == NULL) {
            return true;
        }
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
};

• Version 3.0

/**
 * 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:
    bool hasPathSum(TreeNode* root, int sum) {
        return root && ((root->val == sum && !root->left && !root->right) || (hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val)));
    }
};

2.2 Iterative

/**
 * 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:
    bool hasPathSum(TreeNode* root, int sum) {    
        if(root == NULL) {
            return false;
        }
        stack<TreeNode*> nodes;
        stack<int> values;
        nodes.push(root);
        values.push(root->val);
        while(!nodes.empty()) {
            TreeNode* current = nodes.top();
            nodes.pop();
            int value = values.top();
            values.pop();
            if(value == sum && current->left == NULL && current->right == NULL) {
                return true;
            }
            if(current->left) {
                nodes.push(current->left);
                values.push(value + current->left->val);
            }
            if(current->right) {
                nodes.push(current->right);
                values.push(value + current->right->val);
            }
        }
        return false;
    }
};

Reference

1. https://leetcode.com/problems/path-sum/description/