C++ dfs

class Solution {
public:
    bool isok = false;
    void dfs(TreeNode* root,int nowSum,int targetSum){
        if(isok) return;
        if(root == nullptr) return;
        nowSum += root->val;
        if(nowSum == targetSum && root->left == nullptr && root-> right == nullptr) isok = true;
        if(root->left != nullptr) dfs(root->left,nowSum,targetSum);
        if(root->right != nullptr) dfs(root->right,nowSum,targetSum);
    }

    bool hasPathSum(TreeNode* root, int targetSum) {
       dfs(root,0,targetSum);
       return isok;
    }
};

Java dfs1

class Solution {

    boolean isok = false;
    public void dfs(TreeNode root,int targetSum){
        if(isok) return;
        if(root == null) return;
        //当前节点必须是叶子节点！
        if(targetSum == root.val && root.left == null && root.right == null) isok = true;
        dfs(root.left,targetSum-root.val);
        dfs(root.right,targetSum-root.val);
    }
    
    public boolean hasPathSum(TreeNode root, int targetSum) {
       dfs(root,targetSum);
       return isok;
    }
}

java dfs2

只要有任意一条路走得通就返回true

class Solution {

    public boolean hasPathSum(TreeNode root, int targetSum) {
        if(root == null) return false;
        if(root.left == null && root.right == null) return root.val == targetSum;
        if(hasPathSum(root.left,targetSum-root.val)) return true;
        if(hasPathSum(root.right,targetSum-root.val)) return true;
        return false;
    }
}