平衡二叉树

[题目](110. 平衡二叉树)

重点

要使用后序遍历

代码实现

/**
 * 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 == NULL) {
            return 0;
        }
        // 左
        int leftHeight = getHeight(node->left);
        if (leftHeight == -1) {
            return -1;
        }
        // 右
        int rightHeight = getHeight(node->right);
        if (rightHeight == -1) {
            return -1;
        }
        // 中
        return abs(leftHeight - rightHeight) > 1 ? -1 : 1 + max(leftHeight, rightHeight);
    }

    bool isBalanced(TreeNode* root) {
        return getHeight(root) == -1 ? false : true;
    }
};

二叉树的所有路径

[题目](257. 二叉树的所有路径)

重点

使用前序遍历，因为只有前序遍历是父节点指向子节点
要用到回溯

代码实现

/**
 * 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 traversal(TreeNode* cur, vector<int> & path, vector<string> & result) {
        // 中，因为最后一个节点也要加入到path中
        path.push_back(cur->val);
        // 这才到了叶子节点
        if (cur->left == NULL && cur->right == NULL) {
            string sPath;
            for (int i = 0; i < path.size() - 1; i++) {
                sPath += to_string(path[i]);
                sPath += "->";
            }
            sPath += to_string(path[path.size() - 1]);
            result.push_back(sPath);
            return;
        }
        // 左
        if (cur->left) {
            traversal(cur->left, path, result);
            // 回溯
            path.pop_back();
        }
        // 右
        if (cur->right) {
            traversal(cur->right, path, result);
            // 回溯
            path.pop_back();
        }
    }

    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> result;
        vector<int> path;
        if (root == NULL) {
            return result;
        }
        traversal(root, path, result);
        return result;
    }
};