#104 Maximum Depth of Binary Tree
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
求二叉樹的深度。二叉樹的操作許多都可以用遞歸
//7ms
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
int maxDepth(struct TreeNode* root) {
int l_depth,r_depth;
if(!root)
return 0;
else
{
l_depth = maxDepth(root->left);
r_depth = maxDepth(root->right);
return (l_depth >= r_depth) ? (l_depth+1):(r_depth+1);
}
}
#110 Balanced Binary Tree
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
判斷給定二叉樹是否是平衡二叉樹。。。深度差小於等於1.可以遞歸求每個結點的深度,判斷其子樹高度差。
//8ms
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
int maxDepth(struct TreeNode* root) {
int l_depth,r_depth;
if(!root)
return 0;
else
{
l_depth = maxDepth(root->left);
r_depth = maxDepth(root->right);
return (l_depth >= r_depth) ? (l_depth+1):(r_depth+1);
}
}
bool isBalanced(struct TreeNode* root) {
int k;
if(!root)//空樹
return true;
k=maxDepth(root->left)-maxDepth(root->right);
if(abs(k)>1)
return false;
else
return isBalanced(root->left) && isBalanced(root->right);
}
#111 Minimum Depth of Binary Tree
Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
求二叉樹的最小深度---//4ms
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
int minDepth(struct TreeNode* root) {
int l_depth,r_depth;
if(!root)
return 0;
if(!root->left)
return 1+minDepth(root->right);
if(!root->right)
return 1+minDepth(root->left);
if(root->right && root->left)
{
l_depth = minDepth(root->left);
r_depth = minDepth(root->right);
return (l_depth < r_depth) ? (l_depth+1) : (r_depth+1);
}
}
#112 Path Sum
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:Given the below binary tree and
sum
= 22
,
5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2
which sum is 22.
判定指定二叉樹是否存在路徑和等於指定值。
//8ms
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool hasPathSum(struct TreeNode* root, int sum) {
if(!root)
return false;
if( root->left==NULL && root->right==NULL)
return sum == root->val;
else
return hasPathSum(root->left,sum - root->val) || hasPathSum(root->right,sum - root->val);
}