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【LeetCode】Path Sum
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.
这题就是深度优先遍历(DFS)，使用变量pathSum记录在栈中的节点之和。
栈中的节点就是从根节点到当前节点的路径。
如果当前节点是叶节点，则检查pathSum是否等于sum。
/** * Definition for binary tree * 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; int pathSum = 0; stack<TreeNode*> s; unordered_set<TreeNode*> visited; //visit root s.push(root); pathSum += root->val; visited.insert(root); //whenever add a node into pathSum, check it if(root->left == NULL && root->right == NULL) {//root itself is leaf if(pathSum == sum) return true; } while(!s.empty()) { TreeNode* top = s.top(); if(top->left) {//has left if(visited.find(top->left) == visited.end()) {//not visited //visit s.push(top->left); pathSum += top->left->val; visited.insert(top->left); //judge leaf if(top->left->left == NULL && top->left->right == NULL) {//leaf if(pathSum == sum) return true; } continue; } } if(top->right) {//has right if(visited.find(top->right) == visited.end()) {//not visited //visit s.push(top->right); pathSum += top->right->val; visited.insert(top->right); //judge leaf if(top->right->left == NULL && top->right->right == NULL) {//leaf if(pathSum == sum) return true; } continue; } } s.pop(); //no need to go down, pop pathSum -= top->val; } return false; }};
【LeetCode】Path Sum
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