Path Sum

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Notes:不太会用递归啊

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Integer> list = new ArrayList<Integer>();    public void getLength(TreeNode root, int treeLength){        if(root.left == null && root.right == null){            list.add(treeLength + root.val);        }        if(root.left != null){            getLength(root.left, treeLength + root.val);        }        if(root.right != null){            getLength(root.right, treeLength + root.val);        }    }    public boolean hasPathSum(TreeNode root, int sum) {        int treeLength = 0;        if(root == null){            return false;         }                getLength(root, treeLength);                for(int i = 0; i < list.size(); i++)          {              if(list.get(i) == sum){                return true;            }          }        return false;    }}

简单的递归方法：

/** * 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) {        int treeLength = 0;        if(root == null){            return false;         }        if(root.left == null && root.right == null) return root.val == sum;        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);    }}

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