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.
Solution:
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean hasPathSum(TreeNode root, int sum) { if (root==null) return false; boolean res = hasPathSumRecur(root,sum); return res; } public boolean hasPathSumRecur(TreeNode curNode, int residual){ if (curNode.left==null&&curNode.right==null){ if (residual==curNode.val) return true; else return false; } boolean left = false; boolean right = false; if (curNode.left!=null) left = hasPathSumRecur(curNode.left,residual-curNode.val); if (curNode.right!=null) right = hasPathSumRecur(curNode.right,residual-curNode.val); if (left||right) return true; else return false; } }
This is a recursive problem. No hard.
原文：http://www.cnblogs.com/lishiblog/p/4083040.html