35. Binary Tree Level Order Traversal && Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal

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

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

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

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]
思想： 若递归，传入层号。若迭代，使用队列，在每层结束时，加入一个标记。
方法一：递归：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
void levelPath(TreeNode* root, int level, vector<vector<int> > &path) {
    if(root == NULL) return;
    level < path.size() ? path[level].push_back(root->val) : path.push_back(vector<int>(1, root->val));
    levelPath(root->left, level+1, path);
    levelPath(root->right, level+1, path);
}
class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int> > path;
        levelPath(root, 0, path);
        return path;
    }
};

方法二：迭代

/**
 * 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<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int> > vec;
        if(root == NULL) return vec;
        queue<TreeNode*> qu;
        qu.push(root);
        qu.push(0);
        vector<int> vec2;
        while(!qu.empty()) {
            TreeNode *p = qu.front();
            qu.pop();
            if(!p) {
                if(vec2.size()) { vec.push_back(vec2); vec2.clear();}
                if(!qu.empty()) qu.push(0);
            }else {
                vec2.push_back(p->val);
                if(p->left)   qu.push(p->left);
                if(p->right)  qu.push(p->right);
            }
        }
        return vec;
    }
};

Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

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

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

思想： 目前用两种方法：1 同上，最后将结果反转一下。 2.先求出最大层数，再层序遍历。（也许还有更好的方法）
1.

void levelPath(TreeNode* root, int level, vector<vector<int> > &path) {
    if(root == NULL) return;
    level < path.size() ? path[level].push_back(root->val) : path.push_back(vector<int>(1, root->val));
    levelPath(root->left, level+1, path);
    levelPath(root->right, level+1, path);
}
class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        vector<vector<int> > path;
        levelPath(root, 0, path);
        return vector<vector<int> > (path.rbegin(), path.rend());
    }
};

2.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
int getLevel(TreeNode *root) {
    if(root == NULL) return -1;
    return max(getLevel(root->left), getLevel(root->right)) + 1;
}
void getLevel2(TreeNode *root, int curL, vector<vector<int> > &vec) {
    if(root == NULL) return;
    vec[curL].push_back(root->val);
    getLevelBottom(root->left, curL-1, vec);
    getLevelBottom(root->right,curL-1, vec);
}
class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        int L = getLevel(root);
        vector<vector<int> > vec(L+1, vector<int>());
        getLevelBottom(root, L, vec);
        return vec;
    }
};