37. Binary Tree Zigzag Level Order Traversal && Binary Tree Inorder Traversal

Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes‘ values. (ie, from left to right, then right to left for the next level and alternate between).

For example: Given binary tree {3,9,20,#,#,15,7},

    3
   /   9  20
    /     15   7

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

思路： 使用两个队列（一个可以顺序读，所以用vector模拟），每个队列放一层结点。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
void getQue1(vector<TreeNode*> &q1, queue<TreeNode*> &q2, vector<vector<int> > &vec) {
    while(!q2.empty()) {
        TreeNode *p = q2.front();
        q2.pop();
        if(p->left) q1.push_back(p->left);
        if(p->right) q1.push_back(p->right);
    }
    if(q1.size() == 0) return;
    vector<int> vec2;
    for(int i = q1.size()-1; i >= 0; --i)
        vec2.push_back(q1[i]->val);
    vec.push_back(vec2);
}
void getQue2(queue<TreeNode*> &q2, vector<TreeNode*> &q1, vector<vector<int> > &vec) {
    if(q1.size() == 0) return;
    vector<int> vec2;
    for(int i = 0; i < q1.size(); ++i) {
        if(q1[i]->left) { q2.push(q1[i]->left); vec2.push_back(q1[i]->left->val); }
        if(q1[i]->right) { q2.push(q1[i]->right); vec2.push_back(q1[i]->right->val); }
    }
    if(vec2.size()) vec.push_back(vec2);
    q1.clear();
}
class Solution {
public:
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        vector<vector<int> > vec;
        if(root == NULL) return vec;
        queue<TreeNode*> q2;
        vector<TreeNode*> q1;
        q2.push(root);
        vec.push_back(vector<int>(1, root->val));
        while(!q2.empty()) {
            getQue1(q1, q2, vec);
            getQue2(q2, q1, vec);
        }
        return vec;
    }
};

Binary Tree Inorder Traversal

OJ: https://oj.leetcode.com/problems/binary-tree-inorder-traversal/

Given a binary tree, return the inorder traversal of its nodes‘ values.

For example: Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

题解： 两种方法： 1. 使用栈:  O(n) Time, O(n) Space。 2. Morris traversal (构造线索树), O(n) Time, O(1) Space.

1. 使用栈

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> vec;
        if(root == NULL) return vec;
        TreeNode *p = root;
        stack<TreeNode *> st;
        st.push(p);
        while(p->left) { p = p->left; st.push(p); }
        while(!st.empty()) {
            TreeNode *q = st.top();
            st.pop();
            vec.push_back(q->val);
            if(q->right) {
                q = q->right; st.push(q);
                while(q->left) { q = q->left; st.push(q); }
            }
        }
        return vec;
    }
};

2. Morris Traversal

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
	vector<int> inorderTraversal(TreeNode *root) {
		vector<int> vec;
		TreeNode *cur, *pre;
		cur = root;
		while(cur) {
			if(cur->left == NULL) {
				vec.push_back(cur->val);
				cur = cur->right;
			} else {
				pre = cur->left;
				while(pre->right && pre->right != cur) pre = pre->right;
				if(pre->right == NULL) {
					pre->right = cur;
					cur = cur->left;
				} else {
					pre->right = NULL;
					vec.push_back(cur->val);
					cur = cur->right;
				}
			}
		}
		return vec;
	}
};