题目

给定二叉搜索树的根结点 root，返回值位于范围 [low, high] 之间的所有结点的值的和。

解法一

思路

逐个遍历，满足就累加

代码

/**
 * 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 {

    private int sum = 0;
    public int rangeSumBST(TreeNode root, int low, int high) {
        dfs(root, low, high);
        return sum;
    }

    private void dfs(TreeNode root, int low, int high) {
        if (root == null) {
            return;
        }
        
        dfs(root.left, low, high);
        if (root.val >= low && root.val <= high) {
            sum += root.val;
        }
        dfs(root.right, low, high);
    }
}

解法二

思路

left + (root满足范围) + right

代码

/**
 * 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 int rangeSumBST(TreeNode root, int low, int high) {
        if (root == null) {
            return 0;
        }
        int left = rangeSumBST(root.left, low, high);
        int right = rangeSumBST(root.right, low, high);
        int result = left + right;
        if (root.val >= low && root.val <= high) {
            result += root.val;
        }
        return result;
    }
}

解法三

思路

• BFS

代码

/**
 * 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 int rangeSumBST(TreeNode root, int low, int high) {
        if (root == null) {
            return 0;
        }
        int res = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                if (node.val >= low && node.val <= high) {
                    res += node.val;
                }
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
        }
        return res;
    }
}

上面的实现方式都没有考虑 二叉搜索树 这个特征

---

解法一

思路

• dfs

代码

class Solution {
    public int rangeSumBST(TreeNode root, int low, int high) {
        if (root == null) {
            return 0;
        }
        if (root.val > high) {
            return rangeSumBST(root.left, low, high);
        }
        if (root.val < low) {
            return rangeSumBST(root.right, low, high);
        }
        return root.val + rangeSumBST(root.left, low, high) + rangeSumBST(root.right, low, high);
    }
}

解法二

思路

• bfs

代码

class Solution {
    public int rangeSumBST(TreeNode root, int low, int high) {
        int res = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if (node == null) {
                continue;
            }
            if (node.val > high) {
                queue.offer(node.left);
            } else if (node.val < low) {
                queue.offer(node.right);
            } else {
                res += node.val;
                queue.offer(node.left);
                queue.offer(node.right);
            }
        }
        return res;
    }
}