leetcode 112. Path Sum DFS深度优先遍历

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.

这道题考察的就是每一条路径的sum，找到target即可。

代码如下：

import java.util.ArrayList;import java.util.List;/*class TreeNode {      int val;      TreeNode left;      TreeNode right;      TreeNode(int x) { val = x; }}*/public class Solution {    List<Integer> res=new ArrayList<Integer>();    public boolean hasPathSum(TreeNode root, int sum)     {        if(root==null)            return false;        else            return bydfs(root,sum);    }    public boolean bydfs(TreeNode root, int sum)    {        if(root!=null)        {            if(root.left==null && root.right==null)                return root.val==sum;            boolean left  = bydfs(root.left, sum-root.val);            boolean right = bydfs(root.right, sum-root.val);            return left || right;        }else             return false;    }}

下面是C++的做法，就是做一个DFS深度优先遍历

代码如下：

#include <iostream>#include <vector>using namespace std;/*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)     {        return bydfs(root, 0, sum);    }    bool bydfs(TreeNode* root, int a, int sum)    {        if (root == NULL)            return false;        else        {            if (root->left == NULL && root->right == NULL)            {                if (a + root->val == sum)                    return true;                else                    return false;            }            else            {                bool left = bydfs(root->left, a + root->val, sum);                bool right = bydfs(root->right, a + root->val, sum);                return left || right;            }        }    }};