[LeetCode]Binary Tree Right Side View

和层有关，显然bfs。记录每一个节点的层数，每次用新加入队列节点的值更新当前层的答案。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
struct Node {
    int layer;
    TreeNode *node;
    Node(int l, TreeNode *nd) : layer(l), node(nd) {};
};

class Solution {
public:
    vector<int> rightSideView(TreeNode *root) {
        vector<int> res;
        queue<Node> que;
        if(root == NULL) return res;

        que.push(Node(0, root));
        while(!que.empty()) {
            Node cur = que.front();
            que.pop();
            if(res.size() > cur.layer)
                res[cur.layer] = cur.node ->val;
            else res.push_back(cur.node ->val);
            if(cur.node ->left != NULL)
                que.push(Node(cur.layer + 1, cur.node ->left));
            if(cur.node ->right != NULL)
                que.push(Node(cur.layer + 1, cur.node ->right));
        }
        return res; // 第一次提交的时候这里忘记返回竟然也通过了..
    }
};

/**

* Definition for binary tree

* struct TreeNode {

* int val;

* TreeNode *left;

* TreeNode *right;

* TreeNode(int x) : val(x), left(NULL), right(NULL) {}

* };

struct Node {

int layer;

TreeNode *node;

Node(int l, TreeNode *nd) : layer(l), node(nd) {};

};

class Solution {

public:

vector<int> rightSideView(TreeNode *root) {

vector<int> res;

queue<Node> que;

if(root == NULL) return res;

que.push(Node(0, root));

while(!que.empty()) {

Node cur = que.front();

que.pop();

if(res.size() > cur.layer)

res[cur.layer] = cur.node ->val;

else res.push_back(cur.node ->val);

if(cur.node ->left != NULL)

que.push(Node(cur.layer + 1, cur.node ->left));

if(cur.node ->right != NULL)

que.push(Node(cur.layer + 1, cur.node ->right));

}

return res; // 第一次提交的时候这里忘记返回竟然也通过了..

}

};

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16 April, 2015

[LeetCode]Binary Tree Right Side View