leetcode 101.对称二叉树 golang实现

描述

给定一个二叉树，检查它是否是镜像对称的。

例如，二叉树[1,2,2,3,4,4,3] 是对称的。

    1
   / \
  2   2
 / \ / \
3  4 4  3
但是下面这个[1,2,2,null,3,null,3] 则不是镜像对称的:

    1
   / \
  2   2
   \   \
   3    3
说明:

如果你可以运用递归和迭代两种方法解决这个问题，会很加分。

思路

递归
比较节点的左右节点的值即可

非递归
1. 也是相反判断
2. 注意判断top的节点left.Right  right.Left

实现

递归实现
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isSymmetric(root *TreeNode) bool {
    if root == nil{
        return true
    }

    return compare(root.Left, root.Right)
}

func compare(r1 *TreeNode, r2 *TreeNode)bool{
    if r1 == nil && r2 == nil{
        return true
    }else if r1 == nil || r2 == nil{
        return false
    }else if r1.Val != r2.Val{
        return false
    }
    
    b1 := compare(r1.Left, r2.Right)
    b2 := compare(r1.Right, r1.Left)
    
    return b1 && b2
}

// 非递归实现
func isSymmetric(root *TreeNode) bool {
    stackLeft := make([]*TreeNode, 0)
    stackRight := make([]*TreeNode, 0)

    curNodeLeft := root
    curNodeRight := root
    for (curNodeLeft != nil && curNodeRight != nil) || (len(stackLeft) > 0 && len(stackRight) > 0){
        for curNodeLeft != nil && curNodeRight != nil{
            stackLeft = append(stackLeft, curNodeLeft)
            stackRight = append(stackRight, curNodeRight)
            curNodeLeft = curNodeLeft.Left
            curNodeRight = curNodeRight.Right
            if curNodeLeft == nil && curNodeRight == nil{
                continue
            }else if curNodeLeft == nil || curNodeRight == nil{
                return false
            }
        }

        if len(stackLeft) != len(stackRight){
            return false
        }

        topLeft := stackLeft[len(stackLeft) - 1]
        topRight := stackRight[len(stackRight) - 1]
        if topLeft.Val != topRight.Val{
            return false
        }
        stackLeft = stackLeft[0: len(stackLeft) - 1]
        stackRight = stackRight[0: len(stackRight) - 1]

        curNodeLeft = topLeft.Right
        curNodeRight = topRight.Left
        if curNodeLeft == nil && curNodeRight == nil{
            continue
        }else if curNodeLeft == nil || curNodeRight == nil{
            return false
        }
    }

    return true
}