814. 二叉树剪枝

解法一

递归计算子树的结点和，如果和为0则进行剪枝。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode() {}
 * TreeNode(int val) { this.val = val; }
 * TreeNode(int val, TreeNode left, TreeNode right) {
 * this.val = val;
 * this.left = left;
 * this.right = right;
 * }
 * }
 */
class Solution {
    public TreeNode pruneTree(TreeNode root) {
        if (sumOfSubTree(root) == 0) {
            return null;
        } else {
            return root;
        }
    }
    private int sumOfSubTree(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int sum = root.val;
        int leftSum;
        int rightSum;
        if (root.left != null) {
            leftSum = sumOfSubTree(root.left);
            if (leftSum == 0) {
                root.left = null;
            }
            sum += leftSum;
        }
        if (root.right != null) {
            rightSum = sumOfSubTree(root.right);
            if (rightSum == 0) {
                root.right = null;
            }
            sum += rightSum;
        }
        if (sum == 0) {
            root = null;
        }
        return sum;
    }
}

解法二

评论区看到的解法，比我第一次自己写的简洁多了。不从求和角度入手来判断，递归完成后只需要考虑左右子结点和本身即可。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode() {}
 * TreeNode(int val) { this.val = val; }
 * TreeNode(int val, TreeNode left, TreeNode right) {
 * this.val = val;
 * this.left = left;
 * this.right = right;
 * }
 * }
 */
class Solution {
    public TreeNode pruneTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        root.left = pruneTree(root.left);
        root.right = pruneTree(root.right);
        if ((root.left == null) && (root.right == null) && (root.val == 0)) {
            root = null;
        }
        return root;
    }
}