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); } }