226. Invert Binary Tree

本文介绍了一种在白板上实现反转二叉树的方法,并提供了详细的Java代码示例。该算法通过递归方式交换每个节点的左右子树来完成整个二叉树的反转。

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以下博客纯属上手,大牛们勿喷。
**问题描述:
Invert a binary tree.
4
/ \
2 7
/ \ / \
1 3 6 9
to
4
/ \
7 2
/ \ / \
9 6 3 1**

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.
谷歌:虽然我们 90% 工程师都在用你写的软件(Homebrew),但你不能在白板上反转二叉树,所以滚蛋。

下面附上代码:
/

**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root==null){
            return null;
        }
        TreeNode temp = root.left;
        root.left = root.right;
        root.right  = temp;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }
}
import cv2 import numpy as np def is_approx_rect(contour, epsilon_factor=0.02): peri = cv2.arcLength(contour, True) approx = cv2.approxPolyDP(contour, epsilon_factor * peri, True) return (4 <= len(approx) <= 5 and cv2.isContourConvex(approx)), approx def calc_center(approx): M = cv2.moments(approx) if M["m00"] == 0: return None return int(M["m10"] / M["m00"]), int(M["m01"] / M["m00"]) def distance(p1, p2): return np.sqrt((p1[0]-p2[0])**2 + (p1[1]-p2[1])**2) def main(): cap = cv2.VideoCapture("222.mp4") if not cap.isOpened(): print("打开视频失败") return prev_center = None while True: ret, frame = cap.read() if not ret: break gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY) _, binary = cv2.threshold(gray, 120, 255, cv2.THRESH_BINARY_INV) closed = cv2.morphologyEx(binary, cv2.MORPH_CLOSE, cv2.getStructuringElement(cv2.MORPH_RECT, (50, 50))) contours_data = cv2.findContours(closed, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE) contours = contours_data[1] if len(contours_data) == 3 else contours_data[0] candidates = [] for cnt in contours: is_rect, approx = is_approx_rect(cnt) if is_rect: center = calc_center(approx) if center: candidates.append((approx, center, cv2.contourArea(approx))) if not candidates: selected = None elif prev_center is None: selected = max(candidates, key=lambda x: x[2]) else: candidates.sort(key=lambda x: distance(x[1], prev_center)) top_n = [candidates[0]] for c in candidates[1:]: if distance(c[1], prev_center) - distance(candidates[0][1], prev_center) < 50: top_n.append(c) else: break selected = max(top_n, key=lambda x: x[2]) display_frame 将上述代码改成适用于 openmv4 h7 plus 的代码要求给出完整代码
最新发布
08-03
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