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.
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/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public
class Solution { /** * This problem can be solved either by dfs or bfs.<br> * @param root * @param sum * @return */ public
boolean hasPathSum(TreeNode root, int
sum) { boolean
exists = false; if(root != null){ Map<TreeNode, Integer> sumAtNode = new
HashMap<TreeNode, Integer>(); sumAtNode.put(root, root.val); Deque<TreeNode> nodes = new
LinkedList<TreeNode>(); nodes.add(root); while(nodes.peek()!= null){ TreeNode aNode = nodes.poll(); if(aNode.left == null
&& aNode.right == null){ if(sumAtNode.get(aNode) == sum){ exists = true; break; } } else{ int
pareValue = sumAtNode.get(aNode); if(aNode.left != null){ sumAtNode.put(aNode.left, aNode.left.val + pareValue); nodes.add(aNode.left); } if(aNode.right != null){ sumAtNode.put(aNode.right, aNode.right.val + pareValue); nodes.add(aNode.right); } } } } return
exists; }} |
leetcode--Path Sum,布布扣,bubuko.com
原文地址：http://www.cnblogs.com/averillzheng/p/3773952.html
