生命的意义

在于活出自我

而不是成为别人眼中的你

                                —— 24.11.3

938. 二叉搜索树的范围和

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。

示例 1：

输入：root = [10,5,15,3,7,null,18], low = 7, high = 15
输出：32

示例 2：

输入：root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10
输出：23

提示：

• 树中节点数目在范围 [1, 2 * 104] 内
• 1 <= Node.val <= 105
• 1 <= low <= high <= 105
• 所有 Node.val 互不相同

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方法1 中序遍历判断范围

思路

由题目给出的输入序列可以看出，输入序列按照二叉树的层次遍历顺序进行排序

我们按照中序遍历（左 - 根 - 右）的顺序排列给出的节点，遍历树节点，如果树节点在此范围内，则将该节点的值加入总和中，若不在该范围内，则跳过该节点，观察下一个节点，直到遍历完所有节点为止

返回总和

/**
 * 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 node,int low,int high){
        TreeNode p = node;
        LinkedList<TreeNode> stack = new LinkedList<>();
        int sum = 0;
        while(p != null || !stack.isEmpty()){
            if (p != null || !stack.isEmpty()){
                if (p != null){
                    stack.push(p);
                    // 左
                    p = p.left;
                }else {
                    TreeNode pop = stack.pop();
                    // 处理值
                    if (pop.val >= low && pop.val <= high){
                        sum += pop.val;
                    }
                    // 右
                    p = pop.right;
                }
            }
        }
        return sum;
    }
}

 

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方法2 中序遍历剪枝优化 

由于二叉搜索树的特性（左 < 根 < 右），当我们遍历到某一结点时发现该节点不在范围内，则其左子树/右子树可以直接进行剪枝操作

/**
 * 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 node,int low,int high){
        TreeNode p = node;
        LinkedList<TreeNode> stack = new LinkedList<>();
        int sum = 0;
        while(p != null || !stack.isEmpty()){
            if (p != null || !stack.isEmpty()){
                if (p != null){
                    stack.push(p);
                    // 左
                    p = p.left;
                }else {
                    TreeNode pop = stack.pop();
                    // 处理值
                    if (pop.val > high){
                        break;
                    }
                    if (pop.val >= low){
                        sum += pop.val;
                    }
                    // 右
                    p = pop.right;
                }
            }
        }
        return sum;
    }
}

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方法3 上下限递归剪枝优化

思路

从根节点开始进行递归，对于每个节点进行判断，若它的val值大于要求的范围，则将其右子树跳过不用遍历判断，若其val值小于要求的范围，则将其左子树跳过不用遍历判断，如此可大大提高程序的效率 （通过范围上下限判定进行剪枝操作）

/**
 * 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 node,int low,int high){
        if (node == null){
            return 0;
        }
        if (node.val < low){
            // 将节点的左子树省去
            return rangeSumBST(node.right,low,high);
        }
        if (node.val > high){
            // 将节点的右子树省去
            return rangeSumBST(node.left,low,high);
        }

        return node.val + rangeSumBST(node.left,low,high) + rangeSumBST(node.right,low,high);
    }
}