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  • 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.

    递归实现

    从树根开始判断

    • 如果当前是叶子节点等于剩余和,那么程序返回真
    • 如果当前节点为空,则放回false
    • 除了以上情况,返回递归调用左右子节点的结果或值,并且传入剩余和扣除当前节点值后的值
    class Solution {
    public:
        bool hasPathSum(TreeNode *root, int sum) {
            if(root == NULL) return false;
            if(root->left == NULL && root->right == NULL && sum == root->val) return true;
            else return hasPathSum(root->left,sum-root->val) || hasPathSum(root->right, sum-root->val);
        }
    };
    class Solution {
    public:
        bool hasPathSum(TreeNode *root, int sum) {
            if(root == NULL) return false;
            if(root->left == NULL && root->right == NULL) return sum == root->val;
            sum-=root->val;
            return hasPathSum(root->left,sum) || hasPathSum(root->right, sum);
        }
    };
    递归实现

     非递归遍历(根据层次遍历实现即可)

    /**
     * Definition for binary tree
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    
    #include <iostream>
    #include <queue>
    
    using namespace std;
    
    struct TreeNode{
        int val;
        TreeNode *left;
        TreeNode *right;
        TreeNode(int x): val(x), left(NULL),right(NULL){}
    };
    
    struct TreeNodeSum{
        TreeNode *node;
        int sum;
        TreeNodeSum(TreeNode* node_, int sum_ ):node(node_),sum(sum_){}
    };
    
    bool hasPathSum(TreeNode *root, int sum) {
        if(root == NULL) return false;
        queue<TreeNodeSum *> que;
        que.push(new TreeNodeSum(root,root->val));
        int res = 0;
        while(!que.empty()){
            TreeNodeSum *tmp = que.front();que.pop();
            TreeNode *node = tmp->node;
            cout<<node->val<<endl;
            if(!node->left && !node->right && tmp->sum == sum) return true; 
            if(node->left){
                TreeNodeSum *nodeSum = new TreeNodeSum(node->left,node->left->val+tmp->sum);
                que.push(nodeSum);
            }
            if(node->right ){
                TreeNodeSum *nodeSum = new TreeNodeSum(node->right,node->right->val+tmp->sum);
                que.push(nodeSum);
            }
        }
        return false;
    }
    
    int main(){
        TreeNode *root = new TreeNode(1);
    
        TreeNode *root1 = new TreeNode(-2);
    
        TreeNode *root2 = new TreeNode(-3);
    
        TreeNode *root3 = new TreeNode(1);
    
        TreeNode *root4 = new TreeNode(3);
        TreeNode *root5 = new TreeNode(-2);
        TreeNode *root6 = new TreeNode(-1);
        root->left = root1;
        root->right = root2;
        root1->left = root3;
        root1->right = root4;
        root2->left = root5;
        root3->left = root6;
        cout<<hasPathSum(root,-1)<<endl;
    }
    非递归实现
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  • 原文地址:https://www.cnblogs.com/xiongqiangcs/p/3818959.html
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