簡單題

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題目 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);} }

?

轉(zhuǎn)載于:https://www.cnblogs.com/jiajiaxingxing/p/4565053.html

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