标签：
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.
想了半天也没想明白，只好看答案，拍案叫绝。
方法一： recursion. 弱点依然是recursion.
方法二: use queue, 用BFS。
Add all node to a queue and store sum value of each node to another queue. When it is a leaf node, check the stored sum value.
For the tree above, the queue would be: 5 - 4 - 8 - 11 - 13 - 4 - 7 - 2 - 1. It will check node 13, 7, 2 and 1. This is a typical breadth first search(BFS) problem.
记住！！！！！
Java code:
方法一： recursion. 弱点依然是recursion.
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); }
方法二: use queue, 用BFS。
/** * 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; LinkedList<TreeNode> nodes = new LinkedList<TreeNode>(); LinkedList<Integer> values = new LinkedList<Integer>(); nodes.add(root); values.add(root.val); while(!nodes.isEmpty()){ TreeNode curr = nodes.poll(); int sumValue = values.poll(); if(curr.left == null && curr.right == null && sumValue==sum){ return true; } if(curr.left != null){ nodes.add(curr.left); values.add(sumValue+curr.left.val); } if(curr.right != null){ nodes.add(curr.right); values.add(sumValue+curr.right.val); } } return false; } }
Reference:
1. http://www.programcreek.com/2013/01/leetcode-path-sum/
标签：
原文地址：http://www.cnblogs.com/anne-vista/p/4815024.html
