9.16 基本树遍历

--- knowledge ---
# of leaves = # of nonleaves + 1

- Binary Tree Inorder Traversal

The point is to be done visiting all nodes in left subtree before procceding self then right tree.
Visit self when it has no left child. Then do the same traversal on right subtree.

vector<int> inorderTraversal(TreeNode* root) {
        vector<int> ret;
        stack<TreeNode*> s;
        TreeNode* p = root;
        while (!s.empty() || p) {
            while (p) {
                s.push(p);
                p = p->left;
            }
            if (!s.empty()) {
                p = s.top();
                s.pop();
                ret.push_back(p->val);
                p = p->right;
            }   
        }
        return ret;
    }

- Binary Tree Preorder Traversal

    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> ret;
        stack<TreeNode*> s;
        if (root) s.push(root);
        while (!s.empty()) {
            auto p = s.top();
            s.pop();
            if (p->right) s.push(p->right);
            ret.push_back(p->val);
            if (p->left) s.push(p->left);
        }
        return ret;
    }

或者和其他统一，大家都这么写。。？

    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> ret;
        stack<TreeNode*> s;
        TreeNode* p = root;
        while (!s.empty() || p) {
            while (p) {
                ret.push_back(p->val);
                s.push(p);
                p = p->left;
            }
            if (!s.empty()) {
                p = s.top()->right;
                s.pop();
            }
        }
        return ret;
    }

- Binary Tree Postorder Traversal

Essentially for visiting a tree rooted at self, we want to visit left tree, right tree, then self->val.

The idea here is to record self->val the second time it appears at stack top.

Think about how it works: on stack , we push self, right, left

    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> v;
        stack<pair<TreeNode*, bool>> s; //snd is true when occurred on stack top
        if (root) s.push(make_pair(root, false));
        pair<TreeNode*, bool> p;
        while (!s.empty()) {
            p = s.top();
            if (p.second) {
                v.push_back(p.first->val);
                s.pop();
            } else {
                s.top().second = true; // bug before, p.second is another copy, not reference
                if (p.first->right) s.push(make_pair(p.first->right, false));
                if (p.first->left)  s.push(make_pair(p.first->left, false));
            }
        }
        return v;
    }

http://www.cnblogs.com/dolphin0520/archive/2011/08/25/2153720.html

with pic
http://zisong.me/post/suan-fa/geng-jian-dan-de-bian-li-er-cha-shu-de-fang-fa

- 107] Binary Tree Level Order Traversal II

below essentially uses levelNum to keep track of # of nodes in last level,
or can also insert a nullptr to indicate end of each level

public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
        
        if(root == null) return wrapList;
        
        queue.offer(root);
        while(!queue.isEmpty()){
            int levelNum = queue.size();
            List<Integer> subList = new LinkedList<Integer>();
            for(int i=0; i<levelNum; i++) {
                if(queue.peek().left != null) queue.offer(queue.peek().left);
                if(queue.peek().right != null) queue.offer(queue.peek().right);
                subList.add(queue.poll().val);
            }
            wrapList.add(subList);
        }
        return wrapList;
    }
}

level order, reversed : apply to last one also

    void dfs(TreeNode* root, vector<vector<int>>& ret, int level) {
        if (!root) return;
        if (level == ret.size()) {
            ret.push_back(vector<int>{});
        }
        ret[level].push_back(root->val);
        if (root->left) dfs(root->left, ret, level+1);
        if (root->right) dfs(root->right, ret, level+1);
    }
    
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> ret{};
        dfs(root, ret, 0);
        return vector<vector<int>>(ret.rbegin(), ret.rend());
    }

104] Maximum Depth of Binary Tree

depth is the max num of nodes from root to any leaf, so nullptr has depth of 0

1) bfs w/ nullptr as marker

    // bfs
    int maxDepth(TreeNode* root) {
        if (!root) return 0;
        queue<TreeNode*> q;
        q.push(root);
        q.push(nullptr);
        
        int dep = 0; // dep of finished level
        while (1) {
            auto curr = q.front();
            q.pop();
            if (!curr) {
                ++dep;
                if (q.empty()) return dep;
                q.push(nullptr);
            } else {
                if (curr->left) q.push(curr->left);
                if (curr->right) q.push(curr->right);
            }
        }
        return -1; 
    }

2) dfs using 2 stacks (THINK which 2 stacks)

    int maxDepth(TreeNode* root) {
        if (!root) return 0;
        stack<TreeNode*> path;
        stack<int> maxDep; 
        path.push(root);
        maxDep.push(1);
        
        int ret = -1;
        while (path.size()) {
            auto curr = path.top();
            path.pop();
            int currDep = maxDep.top();
            maxDep.pop();
            if (curr->left) {
                path.push(curr->left);
                maxDep.push(currDep+1);
            }
            if (curr->right) {
                path.push(curr->right);
                maxDep.push(currDep+1);
            }
            if (!curr->left && !curr->right) ret = max(ret, currDep); // technically.. but can check every time
        }
        return ret;
    }

- 226] Invert Binary Tree

iterative dfs simulating recursion w/ stack

    TreeNode* invertTree(TreeNode* root) {
        stack<TreeNode*> s;
        s.push(root);
        while (s.size()) {
            auto curr = s.top();
            s.pop();
            if (!curr) continue;
            s.push(curr->left);
            s.push(curr->right);
            swap(curr->left, curr->right);
        }
        return root;
    }

- 100] Same Tree

how to simulate rec dfs w/ stack again

    bool isSameTree(TreeNode* p, TreeNode* q) {
        stack<TreeNode*> s1, s2;
        s1.push(p);
        s2.push(q);
        
        while (s1.size() && s2.size()) {
            auto t1 = s1.top();
            auto t2 = s2.top();
            s1.pop();
            s2.pop();
            if ((!t1) != (!t2)) return false;
            if (!t1) continue;
            if (t1->val!=t2->val) return false;
            
            s1.push(t1->left);
            s1.push(t1->right);
            s2.push(t2->left);
            s2.push(t2->right);
        }
        return s1.size()==s2.size();
    }

235] Lowest Common Ancestor of a Binary Search Tree

    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (!root || !p || !q) return nullptr;
        if ((p->val < root->val) && (q->val < root->val))  return lowestCommonAncestor(root->left, p, q);
        if ((p->val > root->val) && (q->val > root->val))  return lowestCommonAncestor(root->right, p, q);
        return root;
    }

iterative
also can use difference b/w p->val and root->val to determine relation. if the multiplication gives >0(loop again), <0, =0

    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (!root || !p || !q) return nullptr;
        while ((!(p->val < root->val)) == (!(q->val < root->val)) ) { // invariant in loop: both in L/ both >=root
            if (p->val==root->val || q->val==root->val) break; // eliminates the equal case
            root = p->val < root->val? root->left : root->right;
        }
        return root;
    }

- 101] Symmetric Tree

same trick, more practice

    bool isSymmetric(TreeNode* root) {
        if (!root) return true;
        stack<TreeNode*> s1, s2;
        s1.push(root->left);
        s2.push(root->right);
        while (s1.size()) {
            auto t1 = s1.top(), t2 = s2.top();
            s1.pop();
            s2.pop();
            if (!t1 && !t2) continue;
            if (!t1 || !t2 || t1->val!=t2->val) return false;
            s1.push(t1->left);
            s2.push(t2->right);
            s1.push(t1->right);
            s2.push(t2->left);
        }
        return true;
    }

bfs using 2 qs, or just use 1 q following rule of 2n nodes nl, nr..

    bool isSymmetric(TreeNode* root) {
        queue<TreeNode*> q; // always store 2n nodes, (nl, nr, ...)
        q.push(root);
        q.push(root);
        while (q.size()) {
            auto t1 = q.front();
            q.pop();
            auto t2 = q.front();
            q.pop();
            if (!t1 && !t2) continue;
            if (!t1 || !t2 || t1->val!=t2->val) return false;
            q.push(t1->left);
            q.push(t2->right);
            q.push(t1->right);
            q.push(t2->left);
        }
        return true;
    }

    void dfs(TreeNode* root, vector<string>& ret, string& path) {
        int oriLen = path.size();
        path += to_string(root->val);
        if (!root->left && !root->right) {
            ret.push_back(path);
        }
        if (root->left) {
            path += "->";
            dfs(root->left, ret, path);
        }
        path.erase(path.begin()+oriLen+1, path.end()); // erase till last char is root->val
        if (root->right) {
            path += "->";
            dfs(root->right, ret, path);
        }
        path.erase(path.begin()+oriLen, path.end());
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> ret;
        if (!root) return ret;
        string path;
        dfs(root, ret, path);
        return ret;
    }

Binary Tree Paths

for string cancatenation, use += (append) for best efficiency

    void dfs(TreeNode* root, vector<string>& ret, string& path) {
        int oriLen = path.size();
        path += to_string(root->val);
        int oriLen2 = path.size();
        if (!root->left && !root->right) {
            ret.push_back(path);
        }
        if (root->left) {
            path += "->";
            dfs(root->left, ret, path);
        }
        path.erase(path.begin()+oriLen2, path.end()); // erase till last char is root->val
        if (root->right) {
            path += "->";
            dfs(root->right, ret, path);
        }
        path.erase(path.begin()+oriLen, path.end());
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> ret;
        if (!root) return ret;
        string path;
        dfs(root, ret, path);
        return ret;
    }

- Minimum Depth of Binary Tree

node the definition of min path: # of nodes along the shortest path from root to LEAF

    int minDepth(TreeNode* root) {
        if (!root) return 0;
        if (!root->left || !root->right) 
            // because one of them is of depth 0 anyways OR l + r + 1
            return 1 + max(minDepth(root->left), minDepth(root->right)); 
        return 1 + min(minDepth(root->left), minDepth(root->right));
    }

- unique binary search tree

??? read how to do q2 with dp: https://discuss.leetcode.com/topic/2940/java-solution-with-dp

    int numTrees(int n) {
        if (n<2) return 1;
        vector<int> uniqueCt(n+1, 0);
        uniqueCt[0] = uniqueCt[1] = 1;
        //uniqueCt[i] stores number of unique BST that stores 1...i
        for (int i=2; i<n+1; ++i) {
            for (int pivot=0; pivot<=i; ++pivot) {
                uniqueCt[i] += uniqueCt[pivot-1] * uniqueCt[i-pivot];
            }
        }
        return uniqueCt[n];
    }

109] Convert Sorted List to Binary Search Tree

左边n/2，右边n-n/2-1

top to bottom, O(logn) call stack space, O(nlogn) recurse logn times, each find mid
bottom up, time O(n); space O(logn) for callstack??

    // generate BST from list using n nodes
    TreeNode* genTree(ListNode*& head, int n) {
        if (!n) return nullptr;
        TreeNode* root = new TreeNode(-1);
        root->left = genTree(head, n/2);
        // assume built left tree, and head now points to current root pos
        root->val = head->val;
        head = head->next;
        root->right = genTree(head, n-n/2-1);
        return root;
    }
    TreeNode* sortedListToBST(ListNode* head) {
        int n=0;
        for (auto curr=head; curr; curr=curr->next) ++n;
        return genTree(head, n);
    }

- 199] Binary Tree Right Side View

preorder traversal, keep update ret v

    void preOrder(TreeNode* root, vector<int>& ret, int level) {
        if (!root) return;
        if (level==ret.size()) {
            ret.push_back(root->val);
        } else {
            ret[level] = root->val;
        }
        preOrder(root->left, ret, level+1);
        preOrder(root->right, ret, level+1);
    }
    vector<int> rightSideView(TreeNode* root) {
        vector<int> ret;
        preOrder(root, ret, 0);
        return ret;
    }

转念一想，直接优先traverse右树，再左。这样直接看到rightview

- 331] Verify Preorder Serialization of a Binary Tree

MARK```

method 1: if at any point we see # # N from top to bottom of stack, trim the tree by replacing the 3 nodes with a #. using string to mock a stack

    bool isValidSerialization(string preorder) {
        if (preorder.empty()) return false;
        preorder += ',';
        string s;
        char bufferc = preorder[0];
        for (int i=1; i<preorder.size(); ++i) {
            if (preorder[i] == ',') {
                s += bufferc=='#'? '#' : 'N';
                while (s.size()>=3 && s[s.size()-1]=='#' && 
                      s[s.size()-2]=='#' && s[s.size()-3]!='#') {
                    s.replace(s.end()-3, s.end(), 1, '#');
                }
            } else {
                bufferc = preorder[i];
            }
        }
        return s=="#";
    }

method 2:

- keep track of number of null nodes can have: should always >=0 
- careful to check condition immediately after decrement by 1 on reading node

    bool isValidSerialization(string preorder) {
        if (preorder.empty()) return false;
        
        int cap = 1;
        char buffer;
        preorder += ',';
        for (int i=0; i<preorder.size(); ++i) {
            if (preorder[i]==',') {
                if (--cap<0) break;
                if (buffer!='#') cap += 2;
            } else {
                buffer = preorder[i];
            }
        }
        
        return cap==0;
    }

method 3:

The solution is based on the intuition that ,
`num of leaves = num of nonleaves + 1`
We should satisfy the above equation in all the process of the pre-order-traverse of the tree ....
**we just need to find the shortest prefix of the serialization sequence satisfying the property above. If such prefix does not exist, then the serialization is definitely invalid; otherwise, the serialization is valid if and only if the prefix is the entire sequence.**
```c++
    bool isValidSerialization(string preorder) {
        if (preorder.empty()) return false;
        int numNull = 0, numNum = 0, i = 0;
        char buffer;
        preorder += ',';
        
        for (; i<preorder.size() && numNull != numNum + 1; ++i) {
            if (preorder[i]==',') {
                if (buffer=='#') ++ numNull;
                else ++ numNum;
            } else {
                buffer = preorder[i];
            }
            
        }
        return i==preorder.size() && numNull == numNum + 1;
    }

- 114] Flatten Binary Tree to Linked List

    TreeNode* findTail(TreeNode* root) {
        while (root && root->right) root = root->right;
        return root;
    }
    
    void flatten(TreeNode* root) {
        if (!root) return;
        flatten(root->right);
        if (root->left) {
            flatten(root->left);
            auto ltail = findTail(root->left);
            auto tmp = root->right;
            root->right = root->left;
            ltail->right = tmp;
            root->left = nullptr;
        }
    }

- 103] Binary Tree Zigzag Level Order Traversal

while dfs, use level to determine whether insert from begin/end

    // levels starts with 0, and we reverse on odd level 
    void dfs(TreeNode* root, vector<vector<int>>& v, int level) {
        if (!root) return;
        if (level == v.size()) v.push_back(vector<int>{});
        // reverse on odd level: add to front
        if (level%2) v[level].insert(v[level].begin(), root->val);
        else v[level].insert(v[level].end(), root->val);
        dfs(root->left, v, level+1);
        dfs(root->right, v, level+1);
    }
    
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> v;
        dfs(root, v, 0);
        return v;
    }

- 106] Construct Binary Tree from Inorder and Postorder Traversal

    TreeNode* build(vector<int>& inorder, int l1, int r1, vector<int>& postorder, int l2, int r2) {
        if (l1>r1 || r1-l1!=r2-l2) return nullptr;
        TreeNode* root = new TreeNode(postorder[r2]);
        cout<<"root: "<<root->val<<" inorder range "<<l1<<"-"<<r2<<"; "<<"postorder range "<<l2<<"-"<<r2<<endl;
        if (l1==r1) return root;
        
        int inorderRoot = l1, ct=0;
        for (; inorderRoot<inorder.size() && inorder[inorderRoot]!=root->val; ++inorderRoot) {
            ++ct;
        }
        // cout<<"==left: "<<"inorder range "<<l1<<"-"<<inorderRoot-1<<"; "<<"postorder range "<<l2<<"-"<<l2+ct-1<<endl;
        root->left = build(inorder, l1, inorderRoot-1, postorder, l2, l2+ct-1);
        // cout<<"==right: "<<"inorder range "<<inorderRoot+1<<"-"<<r1<<"; "<<"postorder range "<<l2+ct<<"-"<<r2-1<<endl;
        root->right = build(inorder, inorderRoot+1, r1, postorder, l2+ct, r2-1);
        return root;
    }
    
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        int n1 = inorder.size();
        if (postorder.size()!=n1 || !n1) return nullptr;
        return build(inorder, 0, n1-1, postorder, 0, n1-1);
    }

iterative DFS & BFS

DFS:

list nodes_to_visit = {root};
while( nodes_to_visit isn't empty ) {
  currentnode = nodes_to_visit.take_first();
  nodes_to_visit.prepend( currentnode.children );
  //do something
}

BFS:

list nodes_to_visit = {root};
while( nodes_to_visit isn't empty ) {
 currentnode = nodes_to_visit.take_first();
 nodes_to_visit.append( currentnode.children );
 //do something
}

最后编辑于：2017.12.04 03:21:14

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