二叉树

---

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

    boolean isCompleteTreeNode(TreeNode root){if(root == null){return false;}Queue<TreeNode> queue = new LinkedList<TreeNode>();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<Integer> preOrder(TreeNode root){Stack<TreeNode> stack = new Stack<TreeNode>();ArrayList<Integer> list = new ArrayList<Integer>();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<Integer> preOrderReverse(TreeNode root){ArrayList<Integer> result = new ArrayList<Integer>();preOrder2(root,result);return result;}void preOrder2(TreeNode root,ArrayList<Integer> 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<Integer> inOrder(TreeNode root){ArrayList<Integer> list = new ArrayList<<Integer>();Stack<TreeNode> stack = new Stack<TreeNode>();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<Integer> postOrder(TreeNode root){ArrayList<Integer> list = new ArrayList<Integer>();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<Integer> stack = new Stack<Integer>();int currentSum = 0;findPath(r, i, stack, currentSum);}void findPath(TreeNode r,int i,Stack<Integer> 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<Integer> result;ArrayList<Integer> searchRange(TreeNode root,int k1,int k2){result = new ArrayList<Integer>();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<k2){searchHelper(root.right,k1,k2);}}

19.二叉树的层次遍历

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

原文地址：https://www.jianshu.com/p/0190985635eb