http://www.geeksforgeeks.org/bottom-view-binary-tree/
Bottom View of a Binary Tree
Given a Binary Tree, we need to print the bottom view from left to right. A node x is there in output if x is the bottommost node at its horizontal distance. Horizontal distance of left child of a node x is equal to horizontal distance of x minus 1, and that of right child is horizontal distance of x plus 1.
Examples:
20 / 8 22 / \ 5 3 25 / \ 10 14
For the above tree the output should be 5, 10, 3, 14, 25.
If there are multiple bottom-most nodes for a horizontal distance from root, then print the later one in level traversal. For example, in the below diagram, 3 and 4 are both the bottom-most nodes at horizontal distance 0, we need to print 4.
20 / 8 22 / \ / 5 3 4 25 / \ 10 14
For the above tree the output should be 5, 10, 4, 14, 25.
解决思路:算出二叉树最左边节点的距离,在算出二叉树最右边节点的距离,可以得出这棵二叉树所有节点的距离范围,如果根节点的水平距离为9,那么上边两个二叉树的距离范围是[-2, 2]。也就是说,输出节点应该有5个。那么怎么算每个节点的水平距离?首先要层次遍历二叉树,根据规则,根节点的左边孩子的水平距离是根节点水平距离减1,根节点右边孩子水平距离是根节点水平距离加1,层次遍历二叉树过程中,就算出了每个节点的水平距离,但是要求输出的水平距离只对应一个节点,所以要留下水平距离值相同的最后一个节点,用map可以做到。
http://blog.csdn.net/zzran/article/details/41981969
#include <stdio.h> #include <stdlib.h> #include <iostream> #include <vector> #include <queue> #include <map> #include <stack> #include <limits.h> using namespace std; void printArray(int *array, int size) { for(int i = 0; i < size; i++) cout << array[i]<< "\t" ; cout << endl; } void printVector(vector<int> array ) { for(int i = 0; i <array.size(); i++) cout << array[i]<< "\t" ; cout << endl; } struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; void preorder(TreeNode * root) { if(root == NULL) return; cout << root->val << "\t" ; preorder(root->left); preorder(root->right); } void inorder(TreeNode * root) { if(root == NULL) return; inorder(root->left); cout << root->val << "\t" ; inorder(root->right); } void postorder(TreeNode * root) { if(root == NULL) return; postorder(root->left); postorder(root->right); cout << root->val << "\t" ; } struct newNode { TreeNode* m_node; int m_idx; newNode(TreeNode* node, int idx) { m_node = node; m_idx = idx; } }; class Solution { public: vector<int> bottomView(TreeNode* root) { queue<newNode* > q1; queue<newNode* > q2; vector<int> res; map<int, int> mapping;// index -- value pair if(root != NULL) { q1.push(new newNode(root, 0)); } int leftMost = 0; int rightMost = 0; while(!q1.empty()) { newNode * p = q1.front(); q1.pop(); mapping[p->m_idx] = p->m_node->val; if(p->m_idx < leftMost) leftMost = p->m_idx; if(p->m_idx > rightMost) rightMost = p->m_idx; if(p->m_node->left) q2.push(new newNode(p->m_node->left, p->m_idx - 1) ); if(p->m_node->right) q2.push(new newNode(p->m_node->right, p->m_idx + 1 )); if(q1.empty() /*&& !q2.empty()*/) { swap(q1, q2); } } for(map<int, int>::iterator it = mapping.begin(); it != mapping.end(); it++) { cout << it->first <<"\t" <<it->second << endl; } for(int i = leftMost ; i <= rightMost ; i++) res.push_back(mapping[i]); return res; } }; int main() { TreeNode node0(4); TreeNode node1(2); TreeNode node2(7); TreeNode node3(1); TreeNode node4(3); TreeNode node5(5); TreeNode node6(8); node0.left = &node1; node0.right= &node2; node1.left = &node3; node1.right= &node4; node2.left = &node5; node2.right= &node6; Solution sl; vector<int> res = sl.bottomView(&node0); printVector(res); cout << endl; return 0; }
另外,top view也可以用这样的方法,不是保留最后一个,而是保留第一次idx的结构,后续的数据不保存。