1、



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,此题比較简单,採用深度优先策略。

代码例如以下:

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:此题也非常easy。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.

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:上述提示能够看出,相当于二叉树的先序遍历,对每个结点,先訪问根结点,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。

代码例如以下:

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:此题非常easy,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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