标签：des style blog http io for ar art div
OJ: https://oj.leetcode.com/problems/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.
思想：先序遍历。注意的是： 当只有一个孩子结点时，深度是此孩子结点深度加 1 .
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int minDepth(TreeNode *root) {
if(root == NULL) return 0;
if(root->left == NULL && root->right == NULL) return 1;
int l = minDepth(root->left);
int r = minDepth(root->right);
return (l && r) ? min(l, r) + 1 : (l+r+1);
}
};
OJ: https://oj.leetcode.com/problems/minimum-depth-of-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.
思想： 先序遍历。既要返回左右子树判断的结果，又要返回左右子树的深度进行再判断。
所以要么返回一个 pair<bool, int>, 要么函数参数增加一个引用来传递返回值。
方法1：返回一个 pair<bool, int>: (更简洁)
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
typedef pair<bool, int> Pair;
Pair judge(TreeNode *root) {
if(root == NULL) return Pair(true, 0);
Pair L = judge(root->left);
Pair R = judge(root->right);
return Pair(L.first && R.first && abs(L.second-R.second) < 2, max(L.second, R.second)+1);
}
class Solution {
public:
bool isBalanced(TreeNode *root) {
return judge(root).first;
}
};
方法二： 增加一个引用
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
bool judge(TreeNode *root, int& depth) {
if(root == NULL) { depth = 0; return true; }
int l, r;
if(judge(root->left, l) && judge(root->right, r)) {
depth = 1 + max(l, r);
if(l-r <= 1 && l-r >= -1) return true;
else return false;
}
}
class Solution {
public:
bool isBalanced(TreeNode *root) {
int depth;
return judge(root, depth);
}
};
OJ: https://oj.leetcode.com/problems/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.
注： 此题略水，于此带过。
class Solution {
public:
int maxDepth(TreeNode *root) {
return root ? max(maxDepth(root->left), maxDepth(root->right))+1 : 0;
}
};
33. Minimum Depth of Binary Tree && Balanced Binary Tree
标签：des style blog http io for ar art div
原文地址：http://www.cnblogs.com/liyangguang1988/p/3939860.html
