leetcode -day17 Path Sum I II & Flatten Binary Tree to Linked List & Minimum Depth of Binary Tree
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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.
分析,此题比较简单,采用深度优先策略。
代码如下:
class Solution {public: bool hasPathSum(TreeNode *root, int sum) { hasPath = false; if(root){ int tempSum = 0; pathSumCore(root,tempSum,sum); } return hasPath; } void pathSumCore(TreeNode *root,int tempSum, int sum){ if(hasPath){ return; } tempSum += root->val; if(!root->left&&!root->right){ if(tempSum == sum){ hasPath = true; } return; } if(root->left){ pathSumCore(root->left,tempSum,sum); } if(root->right){ pathSumCore(root->right,tempSum,sum); } } bool hasPath;};
2、Path Sum II
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree and sum = 22
,
5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5]]
分析:此题也很简单,常见题,跟上述不同的是保存路径。
代码:
class Solution {public: vector<vector<int> > pathSum(TreeNode *root, int sum) { if(root){ int tempSum = 0; vector<int> path; pathSumCore(root,path,tempSum,sum); } return pathVec; } void pathSumCore(TreeNode* root,vector<int> path,int tempSum,int sum){ tempSum += root->val; path.push_back(root->val); if(!root->left && !root->right){ if(tempSum == sum){ pathVec.push_back(path); } return; } if(root->left){ pathSumCore(root->left,path,tempSum,sum); } if(root->right){ pathSumCore(root->right,path,tempSum,sum); } } vector<vector<int> > pathVec;};
3、Flatten Binary Tree to Linked List
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1 / \ 2 5 / \ \ 3 4 6
The flattened tree should look like:
1 \ 2 \ 3 \ 4 \ 5 \ 6
click to show hints.
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.
分析:上述提示可以看出,相当于二叉树的先序遍历,对每一个结点,先访问根结点,再先序遍历左子树,串在根结点的右孩子上,再先序遍历原右子树,继续串在右孩子上。
代码如下:
class Solution {public: void flatten(TreeNode *root) { if(root){ preOrder(root); } } TreeNode *preOrder(TreeNode *root){ if(!root->left && !root->right){ return root; } TreeNode *lastNode = NULL; TreeNode *rightNode = root->right; //TreeNode *leftNode = root->left; if(root->left){ root->right = root->left; lastNode = preOrder(root->left); root->left = NULL; lastNode->right = rightNode; } if(rightNode){ lastNode = preOrder(rightNode); } return lastNode; }};
4、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.
分析:此题很简单,递归。
class Solution {public: int minDepth(TreeNode *root) { if(!root){ return 0; } n_minDepth = INT_MAX; int tempDepth = 0; minDepthCore(root,tempDepth); return n_minDepth; } void minDepthCore(TreeNode *root,int tempDepth){ ++tempDepth; if(tempDepth >= n_minDepth){ return; } if(!root->left && !root->right){ if(tempDepth < n_minDepth){ n_minDepth = tempDepth; } return; } if(root->left){ minDepthCore(root->left,tempDepth); } if(root->right){ minDepthCore(root->right,tempDepth); } } int n_minDepth;};
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