199. Binary Tree Right Side View (Tree, Stack)

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,

   1            <---
 /  
2     3         <---
     
  5     4       <---

You should return [1, 3, 4].

以下做法不对,还得考虑左边子树比右边子树长的部分

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> ret;
        TreeNode* current = root;
        while(current){
            ret.push_back(current->val);
            if(current->right) current = current->right;
            else current = current->left;
        }
        return ret;
    }
};

Result:Wrong Answer

Input: [1,2,3,4]
Output: [1,3]
Expected: [1,3,4]

所以,得用stack记录左边的子树

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> ret;
        TreeNode* current = root;
        int depth = 0; //depth so far
        int depthGap;
        stack<TreeNode*> nodeStack;
        stack<int> depthStack;
        
        while(current){
            ret.push_back(current->val);
            depth++;
            if(current->right){
                if(current->left){
                    nodeStack.push(current->left);
                    depthStack.push(depth);
                }
                current = current->right;
            } 
            else{
                current = current->left;
            }
        }
        
        while(!nodeStack.empty()){
            current = nodeStack.top();
            nodeStack.pop();
            depthGap = depth - depthStack.top();
            depthStack.pop();
            while(depthGap){
                depthGap--;
                if(current->right){
                    if(current->left){
                        nodeStack.push(current->left);
                        depthStack.push(depth - depthGap);
                    }
                    current = current->right;
                } 
                else{
                    current = current->left;
                }
                if(!current) break;
            }
            if(!current) continue;
            while(current){
                ret.push_back(current->val);
                depth++;
                if(current->right){
                    if(current->left){
                        nodeStack.push(current->left);
                        depthStack.push(depth);
                    }
                    current = current->right;
                } 
                else{
                    current = current->left;
                }
            }
        }
        return ret;
    }
};
原文地址:https://www.cnblogs.com/qionglouyuyu/p/5051035.html