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 <= 1051 <= low <= high <= 105- 所有
Node.val互不相同
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;}
}
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;}
}
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);}
}
