编程之美：层次遍历二叉树若干问题

1.层次遍历

不多说，很简单

2.按深度从下到到上分层遍历二叉树

用栈保存每层即可

#include <iostream>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;

struct node{
  int value;
  node* left;
  node* right;
  node() :left(NULL), right(NULL){}
};

void printNodeByLevelDowntoUp(node* root){
  if (root == NULL)
    return;
  stack<vector<int> >stk;
  queue<node*> Q;
  Q.push(root);
  Q.push(NULL);
  vector<int> level;
  while (!Q.empty()){
    node* n = Q.front();
    Q.pop();
    if (n){
      level.push_back(n->value);
      if (n->left)
        Q.push(n->left);
      if (n->right)
        Q.push(n->right);
    }
    else{
      stk.push(level);
      level.clear();
      if (!Q.empty())
        Q.push(NULL);
    }
  }
  while (!stk.empty()){
    vector<int> vec = stk.top();
    stk.pop();
    for (int i = 0; i<vec.size(); i++){
      cout << vec[i] << " ";
    }
    cout << endl;
  }
}

void nodeLink(node* nodes, int parent, int lchild, int rchild){
  nodes[parent].value = parent;
  if (lchild != -1){
    nodes[parent].left = &nodes[lchild];
  }
  if (rchild != -1){
    nodes[parent].right = &nodes[rchild];
  }
}

int main()
{
  node nodes[9];

  nodeLink(nodes, 1, 2, 3);
  nodeLink(nodes, 2, 4, 5);
  nodeLink(nodes, 3, -1, 6);
  nodeLink(nodes, 4, -1, -1);
  nodeLink(nodes, 5, 7, 8);
  nodeLink(nodes, 6, -1, -1);
  nodeLink(nodes, 7, -1, -1);
  nodeLink(nodes, 8, -1, -1);

  printNodeByLevelDowntoUp(&nodes[1]);

  return 0;
}

结果：

7 8
4 5 6
2 3
1

3.访问第k层节点（从左到右），根节点为0层

用上面的BFS

#include <iostream>
#include<queue>
#include<stack>
using namespace std;

struct node{
  int value;
  node* left;
  node* right;
  node() :left(NULL), right(NULL){}
};

void printNodeLevelK(node* root,int K){
  if (root == NULL)
    return;
  queue<node*> Q;
  Q.push(root);
  Q.push(NULL);
  int level = 0;
  while (!Q.empty()){
    node* n = Q.front();
    Q.pop();
    if (n){
      if (level == K)
        cout << n->value << " ";
      if (n->left)
        Q.push(n->left);
      if (n->right)
        Q.push(n->right);
    }
    else{
      level++;
      if (!Q.empty()){//否则死循环
        Q.push(NULL);
      }
    }
  }
  cout << endl;
}

void nodeLink(node* nodes, int parent, int lchild, int rchild){
  nodes[parent].value = parent;
  if (lchild != -1){
    nodes[parent].left = &nodes[lchild];
  }
  if (rchild != -1){
    nodes[parent].right = &nodes[rchild];
  }
}

int main()
{
  node nodes[9];

  nodeLink(nodes, 1, 2, 3);
  nodeLink(nodes, 2, 4, 5);
  nodeLink(nodes, 3, -1, 6);
  nodeLink(nodes, 4, -1, -1);
  nodeLink(nodes, 5, 7, 8);
  nodeLink(nodes, 6, -1, -1);
  nodeLink(nodes, 7, -1, -1);
  nodeLink(nodes, 8, -1, -1);

  printNodeLevelK(&nodes[1],2);

  return 0;
}

结果：

4 5 6

也可以用递归，见编程之美P253

4.百度面试题：输出二叉树第 m 层的第 k 个节点值(m, k 均从 0 开始计数)

#include <iostream>
#include<queue>
#include<stack>
using namespace std;

struct node{
  int value;
  node* left;
  node* right;
  node() :left(NULL), right(NULL){}
};

void printNodeLevelMK(node* root,int m,int k){
  if (root == NULL)
    return;
  queue<node*> Q;
  Q.push(root);
  Q.push(NULL);
  int level = 0;
  int kk = 0;
  while (!Q.empty()){
    node* n = Q.front();
    Q.pop();
    if (n){
      if (level == m){
        kk++;
        if (kk - 1 == k)
          cout << n->value << endl;
      }
      if (n->left)
        Q.push(n->left);
      if (n->right)
        Q.push(n->right);
    }
    else{
      level++;
      if (!Q.empty()){//否则死循环
        Q.push(NULL);
      }
    }
  }
}

void nodeLink(node* nodes, int parent, int lchild, int rchild){
  nodes[parent].value = parent;
  if (lchild != -1){
    nodes[parent].left = &nodes[lchild];
  }
  if (rchild != -1){
    nodes[parent].right = &nodes[rchild];
  }
}

int main()
{
  node nodes[9];

  nodeLink(nodes, 1, 2, 3);
  nodeLink(nodes, 2, 4, 5);
  nodeLink(nodes, 3, -1, 6);
  nodeLink(nodes, 4, -1, -1);
  nodeLink(nodes, 5, 7, 8);
  nodeLink(nodes, 6, -1, -1);
  nodeLink(nodes, 7, -1, -1);
  nodeLink(nodes, 8, -1, -1);

  printNodeLevelMK(&nodes[1],2,1);

  return 0;
}

结果5