LeetCode-树专题-94-105-102-236-297-543-124-173

94转载自：

https://blog.youkuaiyun.com/weixin_42130471/article/details/80403114

105转载自：

https://blog.youkuaiyun.com/qq_40163148/article/details/89214880

102转载自：

https://blog.youkuaiyun.com/zhangyumengs/article/details/80414858

236转载自：

https://blog.youkuaiyun.com/qq_38595487/article/details/80256894

297-java版本：

public class Codec {
	public String serialize(TreeNode root) {
        if(root == null) return "#,";
        String s = String.valueOf(root.val) + ",";
        s += serialize(root.left);
        s += serialize(root.right);
        return s;
    }
    int index = 0;
    public TreeNode deserialize(String data) {
        String[] items = data.split(",");
        return helper(items);
    }
    
    TreeNode helper(String[] items){
        String s = items[index++];
        if(s.equals("#")) return null;
        TreeNode root = new TreeNode(Integer.valueOf(s));
        root.left = helper(items);
        root.right = helper(items);
        return root;
    }
}

543：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sum = 0;
    int height(TreeNode* root)           //求出树的高度
    {
        if(!root)
            return 0;
        int a = height(root->left);
        int b = height(root->right);
        sum = max(sum, a + b);
        return a > b? a+1: b+1;
    }

    int diameterOfBinaryTree(TreeNode* root) {
        height(root);
        return sum;
    }
};

124java版:

class Solution {
    
    private int ret = Integer.MIN_VALUE;
    
    public int maxPathSum(TreeNode root) {
        getMax(root);
        return ret;
    }
    
    private int getMax(TreeNode r) {
        if(r == null) return 0;
        int left = Math.max(0, getMax(r.left)); // 如果子树路径和为负则应当置0表示最大路径不包含子树
        int right = Math.max(0, getMax(r.right));
        ret = Math.max(ret, r.val + left + right); // 判断在该节点包含左右子树的路径和是否大于当前最大路径和
        return Math.max(left, right) + r.val;
    }
}

173C++中序遍历，用一个栈实现：

class BSTIterator {
private:
    stack<TreeNode*> s;
    
public:
    BSTIterator(TreeNode* root) {
        find_left(root);  
    }
    
    /** @return the next smallest number */
    int next() {
        TreeNode* tmp = s.top();
        s.pop();
        if(tmp->right!=nullptr)  //如果有右子树，下一个点应该是右子树最左边的点，如果无右子树下一个点应该是其父节点，即栈顶元素
            find_left(tmp->right);
        return tmp->val;
        
    }
    
    /** @return whether we have a next smallest number */
    bool hasNext() {
        if(s.empty())
            return false;
        return true;
        
    }
    
    void find_left(TreeNode* root) //找当前节点最左边的结点并一路压栈
    {
        
        while (root!=nullptr)
        {
            s.push(root);
            root = root->left;
        }
    }
};