二叉树算法实战指南

二叉树基础结构

二叉树的每个节点包含三个部分：数据域、左子节点指针和右子节点指针。用C++定义如下：

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

Python实现通常用类表示：

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

核心遍历方法

递归遍历模板

void traverse(TreeNode* root) {
    if (!root) return;
    // 前序位置
    traverse(root->left);
    // 中序位置  
    traverse(root->right);
    // 后序位置
}

迭代遍历实现 层次遍历（BFS）示例：

from collections import deque
def levelOrder(root):
    if not root: return []
    q = deque([root])
    res = []
    while q:
        level = []
        for _ in range(len(q)):
            node = q.popleft()
            level.append(node.val)
            if node.left: q.append(node.left)
            if node.right: q.append(node.right)
        res.append(level)
    return res

关键操作接口

插入节点

public TreeNode insert(TreeNode root, int val) {
    if (root == null) return new TreeNode(val);
    if (val < root.val) 
        root.left = insert(root.left, val);
    else 
        root.right = insert(root.right, val);
    return root;
}

删除节点 需要考虑三种情况：

• 无子节点：直接删除
• 只有一个子节点：用子节点替换
• 有两个子节点：用右子树最小节点替换

OJ实战技巧

路径总和问题

def hasPathSum(root, target):
    if not root: return False
    if not root.left and not root.right:
        return target == root.val
    return hasPathSum(root.left, target-root.val) or hasPathSum(root.right, target-root.val)

最近公共祖先

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    if (!root || root == p || root == q) return root;
    auto left = lowestCommonAncestor(root->left, p, q);
    auto right = lowestCommonAncestor(root->right, p, q);
    if (left && right) return root;
    return left ? left : right;
}

性能优化策略

• 对于频繁查询的问题，考虑使用哈希表存储节点信息
• 在递归过程中利用返回值传递多个信息时，可以使用结构体或pair
• 对于大规模数据，迭代法通常比递归法更节省内存空间

常见错误排查

• 空指针异常：始终检查节点是否为null
• 无限递归：确保递归终止条件正确
• 状态污染：在回溯算法中注意恢复现场
• 边界条件：特别注意树为空或单节点的情况

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