Path Sum I, II

简单题

I

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.

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

II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

典型dfs题目

public class Solution {
    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        ArrayList<Integer> cl = new ArrayList<Integer>();
        dfs(res, cl, root, sum);
        return res; 
    }
    private void dfs(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> cl,TreeNode n, int s ){
        if(n==null) return;
        cl.add(n.val);
        if(n.left==null && n.right==null && n.val==s){
                res.add(new ArrayList<Integer>(cl)); // if dont rebuild, will make conflict with the last line of cl.remove 哦！
        }else{
            if(n.left!=null)
                dfs(res, cl, n.left,s-n.val);
            if(n.right!=null)
                dfs(res, cl, n.right,s-n.val);
        }
        cl.remove(cl.size()-1);
    }
}