二叉树相关算法

---

• 二叉树定义

type TreeNode struct {
         Val int
         Left *TreeNode
         Right *TreeNode
}

---

• 前序遍历（中序遍历、后续遍历只需要调整append的顺序即可）

func preOrder(root *TreeNode) []int {
    var result []int
    if root != nil {
        result = append(result, root.Val)
        result = append(result, preOrder(root.Left)...)
        result = append(result, preOrder(root.Right)...)
    }
    return result
}

---

• 层序遍历

//dfs解法
func levelOrder(root *TreeNode) [][]int {
    var res [][]int
    var dfs func(root *TreeNode, level int)
    dfs = func(root *TreeNode, level int) {
        if root != nil {
            if len(res) == level {
                res = append(res, []int{})
            }
            res[level] = append(res[level], root.Val)
            dfs(root.Left, level+1)
            dfs(root.Right, level+1)
        }
    }
    dfs(root,0)
    return res
}

//  var test [][]int
//  fmt.Println(test)
//  test = append(test, []int{})
//  fmt.Println(test)
//  结果不一样

//bfs解法
func levelOrder(root *TreeNode) [][]int {
    var res [][]int
    if root == nil{
        return res
    }
    var queue = []*TreeNode{root}
    var level int
    for len(queue) > 0 {
        counter := len(queue)
        res = append(res, []int{})
        for counter > 0 {
            counter--
            if queue[0].Left != nil {
                queue = append(queue, queue[0].Left)
            }
            if queue[0].Right != nil {
                queue = append(queue, queue[0].Right)
            }
            res[level] = append(res[level], queue[0].Val)
            queue = queue[1:]
        }
        level++
    }
    return res
}

---

• 二叉树的最大深度

//递归解法
func maxDepth(root *TreeNode) int {
    if root != nil {
        return int(math.Max(float64(maxDepth(root.Left)+1),float64(maxDepth(root.Right)+1)))
    }
    return 0
}

//bfs解法
func maxDepth(root *TreeNode) int {
    if root == nil {
        return 0
    }
    var queue []*TreeNode
    queue = append(queue, root)
    var level int
    for len(queue) > 0 {
        level++
        counter := len(queue)
        for counter > 0 {
            counter--
            if queue[0].Left != nil {
                queue = append(queue, queue[0].Left)
            }
            if queue[0].Right != nil {
                queue = append(queue, queue[0].Right)
            }
            queue = queue[1:]
        }
    }
    return level
}

//dfs解法

func maxDepth(root *TreeNode) int {
    if root == nil {
        return 0
    }
    maxLevel := 1
    var dfs func(root *TreeNode, level int)
    dfs = func(root *TreeNode, level int) {
        if root != nil {
            if level > maxLevel {
                maxLevel = level
            }
            dfs(root.Left, level+1)
            dfs(root.Right, level+1)
        }
    }
    dfs(root,1)
    return maxLevel
}

---

• 对称二叉树

func isSymmetric(root *TreeNode) bool {
    if root == nil {
        return true
    } else {
        return same(root.Left,root.Right)
    }
}

func same(left *TreeNode,right  *TreeNode) bool {
    if left == nil && right ==nil {
        return true
    } else if left ==nil || right == nil  {
        return false
    }  else if left.Val == right.Val {
        return (same(left.Left,right.Right) && same(right.Left,left.Right))
    } else {
        return false
    }
}

---

• 路径总和

//给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有节点值相加等于目标和。
func hasPathSum(root *TreeNode, sum int) bool {
    if root == nil {
        return false
    }
    if root.Left == nil && root.Right == nil {
        return (sum - root.Val) == 0
    }
    return hasPathSum(root.Left, sum-root.Val) ||  hasPathSum(root.Right, sum-root.Val)
}

---

• 二叉树的最近公共祖先

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root == nil || root == p || root == q {
        return root
    }
    left := lowestCommonAncestor(root.Left, p, q)
    right := lowestCommonAncestor(root.Right, p, q)
    if left == nil {
        return right
    } else if right == nil {
        return left
    } else {
        return root
    }
}