zoukankan      html  css  js  c++  java
  • [LeetCode] 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.

    DFS或者BFS。同时,当本节点是叶子节点才判断和,否则节点为空时返回false

     1 /**
     2  * Definition for binary tree
     3  * struct TreeNode {
     4  *     int val;
     5  *     TreeNode *left;
     6  *     TreeNode *right;
     7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     8  * };
     9  */
    10 class Solution {
    11 public:
    12     bool dfs(TreeNode *node, int sum, int curSum)
    13     {
    14         if (node == NULL)
    15             return false;
    16         
    17         if (node->left == NULL && node->right == NULL)
    18             return curSum + node->val == sum;
    19                
    20         return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val);
    21     }
    22     
    23     bool hasPathSum(TreeNode *root, int sum) {
    24         // Start typing your C/C++ solution below
    25         // DO NOT write int main() function
    26         return dfs(root, sum, 0);
    27     }
    28 };
  • 相关阅读:
    [转]优秀的程序员不会觉得累成狗是一种荣耀
    .NET读写Excel工具Spire.XlS使用(DataExport )
    WPF之Binding深入探讨
    第一个WPF应用程序
    Visio作图
    唯一的重复元素
    Strange Problem O(∩_∩)O~
    数据库知识点①
    HDU 2825 Wireless Password
    POJ 1625 Censored!
  • 原文地址:https://www.cnblogs.com/chkkch/p/2767746.html
Copyright © 2011-2022 走看看