• LeetCode 112. Path Sum


    Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

    For example:
    Given the below binary tree and sum = 22,

                  5
                 / 
                4   8
               /   / 
              11  13  4
             /        
            7    2      1
    

    return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

    这个题目其实就是二叉树的遍历,我采用深度遍历

    /**
     * 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:
        void depthFirstSearch(TreeNode* root, int sum, bool &tag)
        {
            if (root==NULL)return;
            if (root->left)
            {
                depthFirstSearch(root->left, sum - root->val, tag);
                depthFirstSearch(root->right, sum - root->val, tag);
            }
            else
            {
                if (root->right)
                {
                    if (root->right->right == NULL&&root->right->left == NULL)
                    {
                        if (sum - root->val - root->right->val == 0)tag = true;
                        return;
                    }
                    else
                    {
                        depthFirstSearch(root->right->left, sum - root->val - root->right->val, tag);
                        depthFirstSearch(root->right->right, sum - root->val - root->right->val, tag);
                    }
                }
                else
                {
                    if (0 == sum-root->val)tag = true;
                }
            }
        }
        bool hasPathSum(TreeNode* root, int sum) {
            bool tag = false;
            depthFirstSearch(root, sum, tag);
            return tag;
        }
    };
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  • 原文地址:https://www.cnblogs.com/csudanli/p/5815174.html
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