剑指offer--4.重建二叉树

https://segmentfault.com/a/1190000008850005?utm_source=tag-newest#articleHeader11

https://www.cnblogs.com/love-yh/p/7423301.html二叉树种类

1.二叉树的高度

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution(object):
    def maxHeight(self, root):
        if root==None:
            return 0
        leftheight=self.maxHeight(root.left)
        rightheight=self.maxHeight(root.right)
        if leftheight>=rightheight:
            return leftheight+1
        else:
            return rightheight+1

2. 分层打印二叉树

class TreeNode:
    def __init__(self,x):
        self.val  = x
        self.left = None
        self.right = None

class Solution:
    def printBinary(self, root):
        queue = []
        res = []
        if root = None:
            return []
        queue.append(root)
        while queue:
            newnode = queue.pop(0)
            res.append(newnode.val)
            if(newnode.left):
                queue.append(newnode.left)
            if(newnode.right):
                queue.append(newnode.right)
        return res

class Solution:
    # 返回二维列表[[1,2],[4,5]]
    def Print(self, pRoot):
        # write code here
        result = []
        nodeList = []
        if pRoot == None:
            return result
        nodeList.append(pRoot)
        while nodeList:
            result_layer = []
            nextLayernodeList = []
            for node in nodeList:
                result_layer.append(node.val)
                if node.left:
                    nextLayernodeList.append(node.left)
                if node.right:
                    nextLayernodeList.append(node.right)
            nodeList = nextLayernodeList
            result.append(result_layer)
        return result

3. 判断是否为平衡二叉树

class TreeNode():
    def __init__(self,x):
        self.val  = x
        self.left = None
        self.right = None
class Solution:
    def getDeepth(self,Root):
        if Root is None:
            return 0
        nright = self.getDeepth(Root.right)
        nleft = self.getDeepth(Root.left)
        return max(nright,nleft)+1
    def IsBalance_solution(self,pRoot):
        if pRoot is None:
            return True
        right = self.getDeepth(pRoot.right)
        left = self.getDeepth(pRoot.left)
        if abs(right - left) > 1:
            return False
        return self.IsBalance_solution(pRoot.right) and self.IsBalance_solution(pRoot.left)

题目：输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

思路：1. 前序遍历首位是根节点，找到根节点之后，根据中序遍历找到左右子树

2. 把找到的根节点从pre里删除，分别找到1中的左右子树, preleft preright vinleft vinright

　　　3. 然后左右子树带入1中

/* function TreeNode(x) {
    this.val = x;
    this.left = null;
    this.right = null;
} */
function reConstructBinaryTree(pre, vin)
{
    var result = null;
    if(pre.length>1) {
        var root= pre[0];
        var vinrootindex = vin.indexOf(root);
        var vinleft = vin.slice(0,vinrootindex);
        var vinright = vin.slice(vinrootindex+1,vin.length);
        pre.shift();
        var preleft = pre.slice(0,vinleft.length);
        var preright = pre.slice(vinleft.length,pre.length);
        result = {
            val : root,
            left : reConstructBinaryTree(preleft,vinleft),
            right : reConstructBinaryTree(preright,vinright)
        }
    }
    else if(pre.length===1) {
        result={
            val: pre[0],
            left: null,
            right: null
        }
    }

    return result;
}

4. 重建二叉树

-*- coding:utf-8 -*-
# 先序遍历特点：第一个值是根节点
# 中序遍历特点：根节点左边都是左子树，右边都是右子树
# 思路：
# 1.首先根据根节点a将中序遍历划分为两部分，左边为左子树，右边为右子树
# 2.使用递归，左右子树再次；；；
#  最后合成一棵树
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    def Re(self,pre,tin):
        if not pre or not tin:
            return None #要求返回的是二叉树，不是数组
        root = TreeNode(pre.pop(0))
        index = tin.index(root.val)
        root.left = self.Re(pre, tin[:index])
        root.right = self.Re(pre, tin[index+1:])
        return root

5. 二叉树的镜像

#交换所有非页节点的子节点
-*- coding:utf -8 -*-
class TreeNode:
    def __init__(self,x):
        self.value = x
        self.left = None
        self.right = None
class Solution:
    def Mirror(self,root):
        if not root:
            return root
        root.left, root.right = root.right, root.left
        self.Mirror(root.left)
        self.Mirror(root.right)
        return root