【算法leetcode每日一练】二叉搜索树的范围和

• 二叉搜索树的范围和：
• 样例 1：
• 样例 2：
• 提示：
• 分析
• 题解
• • java
  • c
  • c++
  • python
  • go
  • rust
  • typescript
• 原题传送门：https://leetcode-cn.com/problems/range-sum-of-bst/

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。

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

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

• 树中节点数目在范围 [1, 2 * 1 0 4 10^4 104] 内
• 1 <= Node.val <= 1 0 5 10^5 105
• 1 <= low <= high <= 1 0 5 10^5 105
• 所有 Node.val 互不相同

• 面对这道算法题目，二当家的陷入了沉思。
• 遍历二叉搜索树是必然的，递归是最直观的方式。

class Solution {
    public int rangeSumBST(TreeNode root, int low, int high) {
        if (root == null) {
			return 0;
		}
		int val;
		if (root.val >= low
			&& root.val <= high) {
			// 根结点符合条件
			val = root.val;
		} else {
			val = 0;
		}
		int leftVal;
		if (root.val > low) {
			// 左子树可能大于等于low
			leftVal = rangeSumBST(root.left, low, high);
		} else {
			leftVal = 0;
		}
		int rightVal;
		if (root.val < high) {
			// 右子树可能小于等于high
			rightVal = rangeSumBST(root.right, low, high);
		} else {
			rightVal = 0;
		}
		return val + leftVal + rightVal;
    }
}

int rangeSumBST(struct TreeNode* root, int low, int high){
    if (root == NULL) {
        return 0;
    }
    int val;
    if (root->val >= low
        && root->val <= high) {
        // 根结点符合条件
        val = root->val;
    } else {
        val = 0;
    }
    int leftVal;
    if (root->val > low) {
        // 左子树可能大于等于low
        leftVal = rangeSumBST(root->left, low, high);
    } else {
        leftVal = 0;
    }
    int rightVal;
    if (root->val < high) {
        // 右子树可能小于等于high
        rightVal = rangeSumBST(root->right, low, high);
    } else {
        rightVal = 0;
    }
    return val + leftVal + rightVal;
}

class Solution {
public:
    int rangeSumBST(TreeNode* root, int low, int high) {
        if (root == nullptr) {
            return 0;
        }
        int val;
        if (root->val >= low
            && root->val <= high) {
            // 根结点符合条件
            val = root->val;
        } else {
            val = 0;
        }
        int leftVal;
        if (root->val > low) {
            // 左子树节点可能大于等于low
            leftVal = rangeSumBST(root->left, low, high);
        } else {
            leftVal = 0;
        }
        int rightVal;
        if (root->val < high) {
            // 右子树节点可能小于等于high
            rightVal = rangeSumBST(root->right, low, high);
        } else {
            rightVal = 0;
        }
        return val + leftVal + rightVal;
    }
};

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rangeSumBST(self, root: TreeNode, low: int, high: int) -> int:
        if root is None:
            return 0
        val = 0
        if root.val >= low and root.val <= high:
            # 根结点符合条件
            val = root.val
        left_val = 0
        if root.val > low:
            # 左子树可能大于等于low
            left_val = self.rangeSumBST(root.left, low, high)
        right_val = 0
        if root.val < high:
            # 右子树可能小于等于high
            right_val = self.rangeSumBST(root.right, low, high)
        return val + left_val + right_val
        

func rangeSumBST(root *TreeNode, low int, high int) int {
    if root == nil {
		return 0
	}
	val := 0
	if root.Val >= low && root.Val <= high {
		// 根结点符合条件
		val = root.Val
	}
	leftVal := 0
	if root.Val > low {
		// 左子树可能大于等于low
		leftVal = rangeSumBST(root.Left, low, high)
	}
	rightVal := 0
	if root.Val < high {
		// 右子树可能小于等于high
		rightVal = rangeSumBST(root.Right, low, high)
	}
	return val + leftVal + rightVal
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option>>,
//   pub right: Option>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    pub fn range_sum_bst(root: Option>>, low: i32, high: i32) -> i32 {
        match root {
            Some(root) => {
                let mut root = root.borrow_mut();
                let val = if root.val >= low && root.val <= high {
                    root.val
                } else {
                    0
                };
                let leftVal = if root.val > low {
                    Self::range_sum_bst(root.left.take(), low, high)
                } else {
                    0
                };
                let rightVal = if root.val < high {
                    Self::range_sum_bst(root.right.take(), low, high)
                } else {
                    0
                };
                val + leftVal + rightVal
            },
            _ => 0
        }
    }
}

function rangeSumBST(root: TreeNode | null, low: number, high: number): number {
    if (!root) {
        return 0;
    }
    let val = 0;
    if (root.val >= low && root.val <= high) {
        // 根结点符合条件
        val = root.val;
    }
    let leftVal = 0
    if (root.val > low) {
        // 左子树可能大于等于low
        leftVal = rangeSumBST(root.left, low, high);
    }
    let rightVal = 0;
    if (root.val < high) {
        // 右子树可能小于等于high
        rightVal = rangeSumBST(root.right, low, high);
    }
    return val + leftVal + rightVal;
};

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