topic:
Given a binary tree, determine whether it is a height-balanced binary tree.
In this question, a height balanced binary tree is defined as:
The absolute value of the height difference between the left and right subtrees of each node in a binary tree does not exceed 1.
Example 1:
Input: root = [3,9,20,null,null,15,7] Output: true
Example 2:
Input: root = [1,2,2,3,3,null,null,4,4] Output: false
Example 3:
Input: root = [] Output: true
hint:
- The number of nodes in the tree is in the
[0, 5000]range -104 <= Node.val <= 104
answer:
class Solution {
public:
// 返回以该节点为根节点的二叉树的高度，如果不是平衡二叉树了则返回-1
int getHeight(TreeNode* node) {
if (node == NULL) {
return 0;
}
int leftHeight = getHeight(node->left);
if (leftHeight == -1) return -1;
int rightHeight = getHeight(node->right);
if (rightHeight == -1) return -1;
return abs(leftHeight - rightHeight) > 1 ? -1 : 1 + max(leftHeight, rightHeight);
}
bool isBalanced(TreeNode* root) {
return getHeight(root) == -1 ? false : true;
}
};topic:
257. All Paths of a Binary Tree
Given the root of a binary tree root , return all paths from the root to the leaves, in any order .
A leaf node is a node that has no child nodes.
Example 1:
Input: root = [1,2,3,null,5] Output: ["1->2->5","1->3"]
Example 2:
Input: root = [1] Output: ["1"]
hint:
- The number of nodes in the tree is in the
[1, 100]range -100 <= Node.val <= 100
answer:
class Solution {
private:
void traversal(TreeNode* cur, vector<int>& path, vector<string>& result) {
path.push_back(cur->val); // 中，中为什么写在这里，因为最后一个节点也要加入到path中
// 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;
}
if (cur->left) { // 左
traversal(cur->left, path, result);
path.pop_back(); // 回溯
}
if (cur->right) { // 右
traversal(cur->right, path, result);
path.pop_back(); // 回溯
}
}
public:
vector<string> binaryTreePaths(TreeNode* root) {
vector<string> result;
vector<int> path;
if (root == NULL) return result;
traversal(root, path, result);
return result;
}
};topic:
Given the root node of a binary tree root , return the sum of all left leaves.
Example 1:
Input: root = [3,9,20,null,null,15,7] Output: 24 Explanation: In this binary tree, there are two left leaves, 9 and 15, so return 24
Example 2:
Input: root = [1] Output: 0
hint:
- The number of nodes is in
[1, 1000]the range -1000 <= Node.val <= 1000
answer:
class Solution {
public:
int sumOfLeftLeaves(TreeNode* root) {
if (root == NULL) return 0;
if (root->left == NULL && root->right== NULL) return 0;
int leftValue = sumOfLeftLeaves(root->left); // 左
if (root->left && !root->left->left && !root->left->right) { // 左子树就是一个左叶子的情况
leftValue = root->left->val;
}
int rightValue = sumOfLeftLeaves(root->right); // 右
int sum = leftValue + rightValue; // 中
return sum;
}
};
Welcome to like, bookmark, comment, your encouragement is the biggest motivation for my creation! (๑╹◡╹)ﾉ"""
Copyright statement: This article is an original article of CSDN blogger "Dumengjiu", which follows the CC 4.0 BY-SA copyright agreement. For reprinting, please attach the original source link and this statement.
Original link: Dumengjiu's blog_CSDN blog-csdn domain blogger