树03

110. 平衡二叉树

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int getHeight(TreeNode* node)
    {
        if(node == nullptr) return 0;
        int leftHeight = getHeight(node->left);
        if(leftHeight == -1) return -1;
        int rightHeight = getHeight(node->right);
        if(rightHeight == -1) return -1;
        if(abs(leftHeight - rightHeight) > 1)
            return -1;
        else
        {
            return max(leftHeight, rightHeight) + 1;
        }
    }
    bool isBalanced(TreeNode* root) {
        int height = getHeight(root);
        if(height == -1) return false;
        else return true;
    }
};

257. 二叉树的所有路径

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void preOrder(TreeNode* node, vector<string> &result, string s)
    {
        s += to_string(node-> val);
        if (node->left == NULL && node->right == NULL) {
            result.push_back(s);
            return;
        }
        
        if(node->left)
            preOrder(node->left, result, s + "->");
        if(node->right)
            preOrder(node->right, result, s + "->");
        
    }
    vector<string> binaryTreePaths(TreeNode* root) {

        vector<string> result;
        if(root == nullptr) return result;
        string s = "";
        preOrder(root, result, s);
        return result;
    }
};

404. 左叶子之和

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void PreOrder(TreeNode* node, int &sum)
    {
        if(node -> left != nullptr && node->left->left == nullptr && node->left->right == nullptr)
        {
            sum += node->left->val;
            // if(node->right) PreOrder(node->right, sum);
            // return;
            
        }
        if(node->left) PreOrder(node->left, sum);
        if(node->right) PreOrder(node->right, sum);
    }
    int sumOfLeftLeaves(TreeNode* root) {
        int sum = 0;
        if(root) 
        {
            PreOrder(root, sum);
            return sum;
        }
        else return sum;

    }
};

513. 找树左下角的值

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int findBottomLeftValue(TreeNode* root) {
        int result;
        queue<TreeNode*> que;
        if(root == nullptr) return 0;
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            int mSize = size;
            while(size--)
            {
                TreeNode* cur = que.front();
                que.pop();
                if(size == mSize - 1)
                    result = cur->val;
                if(cur->left)
                    que.push(cur->left);
                if(cur->right)
                    que.push(cur->right);
            }
        }
        return result;
    }
};