本文来自公众号程序员小乐(study_tech)

作者：IOExceptioner

编辑：www.jianshu.com/p/0190985635eb

先上二叉树的数据结构：

class TreeNode{    int val;    //左孩子    TreeNode left;    //右孩子    TreeNode right;}

二叉树的题目普遍可以用递归和迭代的方式来解

1. 求二叉树的最大深度

int maxDeath(TreeNode node){if(node==null){        return 0;    }int left = maxDeath(node.left);int right = maxDeath(node.right);    return Math.max(left,right) + 1;}

2. 求二叉树的最小深度

  int getMinDepth(TreeNode root){if(root == null){return 0;        }return getMin(root);    }    int getMin(TreeNode root){if(root == null){return Integer.MAX_VALUE;        }if(root.left == null&&root.right == null){return 1;        }return Math.min(getMin(root.left),getMin(root.right)) + 1;    }

3. 求二叉树中节点的个数

  int numOfTreeNode(TreeNode root){if(root == null){            return 0;        }int left = numOfTreeNode(root.left);int right = numOfTreeNode(root.right);        return left + right + 1;    }

4. 求二叉树中叶子节点的个数

     int numsOfNoChildNode(TreeNode root){if(root == null){return 0;        }if(root.left==null&&root.right==null){return 1;        }return numsOfNodeTreeNode(root.left)+numsOfNodeTreeNode(root.right);    }

5. 求二叉树中第k层节点的个数

      int numsOfkLevelTreeNode(TreeNode root,int k){if(root == null||k<1){return 0;            }if(k==1){return 1;            }int numsLeft = numsOfkLevelTreeNode(root.left,k-1);int numsRight = numsOfkLevelTreeNode(root.right,k-1);return numsLeft + numsRight;        }

6. 判断二叉树是否是平衡二叉树

 boolean isBalanced(TreeNode node){        return maxDeath2(node)!=-1;    }int maxDeath2(TreeNode node){if(node == null){            return 0;        }int left = maxDeath2(node.left);int right = maxDeath2(node.right);if(left==-1||right==-1||Math.abs(left-right)>1){            return -1;        }        return Math.max(left, right) + 1;    }

7.判断二叉树是否是完全二叉树

什么是完全二叉树呢？参见

   boolean isCompleteTreeNode(TreeNode root){if(root == null){            return false;        }        Queue queue = new LinkedList();        queue.add(root);        boolean result = true;        boolean hasNoChild = false;while(!queue.isEmpty()){            TreeNode current = queue.remove();if(hasNoChild){if(current.left!=null||current.right!=null){                    result = false;                    break;                }            }else{if(current.left!=null&&current.right!=null){                    queue.add(current.left);                    queue.add(current.right);                }else if(current.left!=null&&current.right==null){                    queue.add(current.left);                    hasNoChild = true;                }else if(current.left==null&&current.right!=null){                    result = false;                    break;                }else{                    hasNoChild = true;                }            }        }        return result;    }

8. 两个二叉树是否完全相同

    boolean isSameTreeNode(TreeNode t1,TreeNode t2){if(t1==null&&t2==null){return true;        }else if(t1==null||t2==null){return false;        }if(t1.val != t2.val){return false;        }        boolean left = isSameTreeNode(t1.left,t2.left);        boolean right = isSameTreeNode(t1.right,t2.right);return left&&right;    }

9. 两个二叉树是否互为镜像

     boolean isMirror(TreeNode t1,TreeNode t2){if(t1==null&&t2==null){return true;        }if(t1==null||t2==null){return false;        }if(t1.val != t2.val){return false;        }return isMirror(t1.left,t2.right)&&isMirror(t1.right,t2.left);    }

10. 翻转二叉树or镜像二叉树

      TreeNode mirrorTreeNode(TreeNode root){if(root == null){            return null;        }        TreeNode left = mirrorTreeNode(root.left);        TreeNode right = mirrorTreeNode(root.right);        root.left = right;        root.right = left;        return root;    }

11. 求两个二叉树的最低公共祖先节点

    TreeNode getLastCommonParent(TreeNode root,TreeNode t1,TreeNode t2){if(findNode(root.left,t1)){if(findNode(root.right,t2)){return root;            }else{return getLastCommonParent(root.left,t1,t2);            }        }else{if(findNode(root.left,t2)){return root;            }else{return getLastCommonParent(root.right,t1,t2)            }        }    }// 查找节点node是否在当前 二叉树中    boolean findNode(TreeNode root,TreeNode node){if(root == null || node == null){return false;        }if(root == node){return true;        }        boolean found = findNode(root.left,node);if(!found){            found = findNode(root.right,node);        }return found;    }

12. 二叉树的前序遍历

迭代解法

  ArrayList preOrder(TreeNode root){        Stack stack = new Stack();        ArrayList list = new ArrayList();if(root == null){return list;        }stack.push(root);while(!stack.empty()){            TreeNode node = stack.pop();list.add(node.val);if(node.right!=null){stack.push(node.right);            }if(node.left != null){stack.push(node.left);            }        }return list;    }

递归解法

  ArrayList preOrderReverse(TreeNode root){        ArrayList result = new ArrayList();        preOrder2(root,result);return result;    }void preOrder2(TreeNode root,ArrayList result){if(root == null){return;        }        result.add(root.val);        preOrder2(root.left,result);        preOrder2(root.right,result);    }

13. 二叉树的中序遍历

  ArrayList inOrder(TreeNode root){        ArrayList list = new ArrayList<();        Stack stack = new Stack();        TreeNode current = root;while(current != null|| !stack.empty()){while(current != null){stack.add(current);                current = current.left;            }            current = stack.peek();stack.pop();list.add(current.val);            current = current.right;        }return list;    }

14.二叉树的后序遍历

  ArrayList postOrder(TreeNode root){        ArrayList list = new ArrayList();if(root == null){return list;        }list.addAll(postOrder(root.left));list.addAll(postOrder(root.right));list.add(root.val);return list;    }

15.前序遍历和后序遍历构造二叉树

  TreeNode buildTreeNode(int[] preorder,int[] inorder){if(preorder.length!=inorder.length){return null;        }return myBuildTree(inorder,0,inorder.length-1,preorder,0,preorder.length-1);    }TreeNode myBuildTree(int[] inorder,int instart,int inend,int[] preorder,int prestart,int preend){if(instart>inend){return null;        }        TreeNode root = new TreeNode(preorder[prestart]);int position = findPosition(inorder,instart,inend,preorder[start]);        root.left = myBuildTree(inorder,instart,position-1,preorder,prestart+1,prestart+position-instart);        root.right = myBuildTree(inorder,position+1,inend,preorder,position-inend+preend+1,preend);return root;    }int findPosition(int[] arr,int start,int end,int key){int i;for(i = start;i<=end;i++){if(arr[i] == key){return i;            }        }return -1;    }

16.在二叉树中插入节点

   TreeNode insertNode(TreeNode root,TreeNode node){if(root == node){return node;        }        TreeNode tmp = new TreeNode();        tmp = root;        TreeNode last = null;while(tmp!=null){            last = tmp;if(tmp.val>node.val){                tmp = tmp.left;            }else{                tmp = tmp.right;            }        }if(last!=null){if(last.val>node.val){                last.left = node;            }else{                last.right = node;            }        }return root;    }

17.输入一个二叉树和一个整数，打印出二叉树中节点值的和等于输入整数所有的路径

   void findPath(TreeNode r,int i){if(root == null){return;        }        Stack stack = new Stack();int currentSum = 0;        findPath(r, i, stack, currentSum);    }void findPath(TreeNode r,int i,Stack stack,int currentSum){        currentSum+=r.val;stack.push(r.val);if(r.left==null&&r.right==null){if(currentSum==i){for(int path:stack){                    System.out.println(path);                }            }        }if(r.left!=null){            findPath(r.left, i, stack, currentSum);        }if(r.right!=null){            findPath(r.right, i, stack, currentSum);        }stack.pop();    }

18.二叉树的搜索区间

给定两个值 k1 和 k2(k1 < k2)和一个二叉查找树的根节点。找到树中所有值在 k1 到 k2 范围内的节点。即打印所有x (k1 <= x <= k2) 其中 x 是二叉查找树的中的节点值。返回所有升序的节点值。

  ArrayList result;ArrayList searchRange(TreeNode root,int k1,int k2){        result = new ArrayList();        searchHelper(root,k1,k2);return result;    }void searchHelper(TreeNode root,int k1,int k2){if(root == null){return;        }if(root.val>k1){            searchHelper(root.left,k1,k2);        }if(root.val>=k1&&root.val<=k2){            result.add(root.val);        }if(root.val            searchHelper(root.right,k1,k2);        }    }

19.二叉树的层次遍历

  ArrayList> levelOrder(TreeNode root){        ArrayList> result = new ArrayList>();if(root == null){return result;        }        Queue queue = new LinkedList();queue.offer(root);while(!queue.isEmpty()){int size = queue.size();            ArrayList< level = new ArrayList():for(int i = 0;i                 TreeNode node = queue.poll();                level.add(node.val);if(node.left != null){queue.offer(node.left);                }if(node.right != null){queue.offer(node.right);                }            }             result.add(Level);        }return result;    }

20.二叉树内两个节点的最长距离

二叉树中两个节点的最长距离可能有三种情况：
1.左子树的最大深度+右子树的最大深度为二叉树的最长距离
2.左子树中的最长距离即为二叉树的最长距离
3.右子树种的最长距离即为二叉树的最长距离
因此，递归求解即可

 private static class Result{  int maxDistance;  int maxDepth;  public Result() {      }  public Result(int maxDistance, int maxDepth) {  this.maxDistance = maxDistance;  this.maxDepth = maxDepth;      }  }  int getMaxDistance(TreeNode root){return getMaxDistanceResult(root).maxDistance;    }Result getMaxDistanceResult(TreeNode root){if(root == null){            Result empty = new Result(0,-1);return empty;        }        Result lmd = getMaxDistanceResult(root.left);        Result rmd = getMaxDistanceResult(root.right);        Result result = new Result();        result.maxDepth = Math.max(lmd.maxDepth,rmd.maxDepth) + 1;        result.maxDistance = Math.max(lmd.maxDepth + rmd.maxDepth,Math.max(lmd.maxDistance,rmd.maxDistance));return result;    }

21.不同的二叉树

给出 n，问由 1…n 为节点组成的不同的二叉查找树有多少种？

  int numTrees(int n ){int[] counts = new int[n+2];        counts[0] = 1;        counts[1] = 1;for(int i = 2;i<=n;i++){for(int j = 0;j                counts[i] += counts[j] * counts[i-j-1];            }        }return counts[n];    }

22.判断二叉树是否是合法的二叉查找树(BST)

一棵BST定义为：
节点的左子树中的值要严格小于该节点的值。
节点的右子树中的值要严格大于该节点的值。
左右子树也必须是二叉查找树。
一个节点的树也是二叉查找树。

 public int lastVal = Integer.MAX_VALUE;public boolean firstNode = true;public boolean isValidBST(TreeNode root) {// write your code hereif(root==null){return true;        }if(!isValidBST(root.left)){return false;        }if(!firstNode&&lastVal >= root.val){return false;        }        firstNode = false;        lastVal = root.val;if (!isValidBST(root.right)) {return false;        }return true;    }

祝大家考试顺利。

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