【leetcode】112. Path Sum

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.

Tips：在一个给定的树上找出是否包含和为sum的路径（路径指从root到根节点）

我采用的递归方法，root有左子树，就用root.left调用作为参数，再次调用函数。

root有右子树，就用root.right调用作为参数，再次调用函数。

public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null)
            return false;
        boolean has = false;
        int ans = 0;
        has = hasPathSumCore(root, sum, ans);
        return has;
    }

    private boolean hasPathSumCore(TreeNode root, int sum, int ans) {
        // TODO Auto-generated method stub
        boolean has = false;
        if (root == null) {
            System.out.println("null");
            return false;
        }
        ans += root.val;
        System.out.println(ans);
        if (ans == sum && root.left == null && root.right == null) {
            has = true;
        }
        if (root.left != null) {
            boolean has1 = hasPathSumCore(root.left, sum, ans);
            has = has || has1;
        }
        if (root.right != null) {
            boolean has2 = hasPathSumCore(root.right, sum, ans);
            has = has || has2;
        }
        ans -= root.val;

        return has;

    }

之后在leetcode上看到更简单的做法（-_-||）

public boolean hasPathSum2(TreeNode root, int sum) {
            if(root == null) return false;
        
            if(root.left == null && root.right == null && sum - root.val == 0) return true;
        
            return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
        }