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Commit 9966120

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Create Solution 112[CPP]
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‎solution/112.Path Sum/README.md‎

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## 路径总和
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### 问题描述
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给定一个二叉树和一个目标和,判断该树中是否存在根节点到叶子节点的路径,这条路径上所有节点值相加等于目标和。
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说明: 叶子节点是指没有子节点的节点。
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示例:
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给定如下二叉树,以及目标和 sum = 22,
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```
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5
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/ \
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4 8
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/ / \
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11 13 4
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/ \ \
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7 2 1
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```
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返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。
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### 思路
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题目要求有没有路径到**叶子节点**使和等于目标值
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主要考察对叶子节点是否判断准确
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这道题很简单,但是准确率不高,原因是的判断条件不明确,左空右不空返回什么什么,右空左不空返回什么什么,调试一直错
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叶子节点唯一判断就是左右空
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`root->left == NULL && root->right==NULL`
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```CPP
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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bool hasPathSum(TreeNode* root, int sum) {
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if(root == NULL)return false;
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if(root->right == NULL && root->left == NULL && sum == root->val)return true;
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bool leftTrue = hasPathSum(root->left,sum - root->val);
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bool rightTrue = hasPathSum(root->right,sum - root->val);
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return (leftTrue || rightTrue);
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}
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};
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```

‎solution/112.Path Sum/Solution.cpp‎

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

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