算法练习day18

一、二叉树搜索树的最小绝对差

二叉搜索树可以看做是有序数组的问题，中序遍历

递归法 中序遍历

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var getMinimumDifference = function (root) {
    let pre = null
    let result = Infinity
    function dfs(root) {
        if (!root) {
            return
        }
        dfs(root.left)
        if (pre !== null) {
            result = Math.min(result, Math.abs(Number(pre.val) - Number(root.val)))
        }
        pre = root
        dfs(root.right)
    }
    dfs(root)
    return result
};

迭代法

var getMinimumDifference = function (root) {
    let result = Infinity
    let pre = null
    let stack = []
    let cur = root
    while(stack.length || cur) {
        if(cur) {
            stack.push(cur)
            cur = cur.left
        } else {
            let node = stack.pop()
            if(pre !== null) {
                result = Math.min(result, Math.abs(Number(node.val - pre.val)))
            }
            pre = node
            cur = node.right
        }
    }
    return result
};

二、二叉搜索树中的众数

递归法

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var findMode = function(root) {
    let result = []
    let pre = null
    let count = 0
    let maxCount = 0
    function dfs(root) {
        if(!root) {
            return
        }
        dfs(root.left)
        if(pre === null) {
            count = 1
        } else if(pre.val === root.val) {
            count++
        } else {
            count = 1
        }
        pre = root
        if(count === maxCount) {
            result.push(root.val)
        }
        if(count > maxCount) {
            maxCount = count
            result = [root.val]
        }
        dfs(root.right)
    }
    dfs(root)
    return result
};

迭代法

var findMode = function(root) {
    let result = []
    let pre = null
    let count = 0
    let maxCount = 0
    let stack = []
    let cur = root
    while(stack.length || cur) {
        if(cur) {
            stack.push(cur)
            cur = cur.left
        } else {
            let node = stack.pop()
            if(!pre) {
                count = 1
            } else if(pre.val === node.val) {
                count++
            } else {
                count = 1
            }
            pre = node
            if(count === maxCount) {
                result.push(node.val)
            }
            if(count > maxCount) {
                maxCount = count
                result = []
                result.push(node.val)
            }
            cur = node.right
        }
    }
    
    return result
};

三、二叉树的最近公共祖先

递归法，后序遍历

拿到左子树和右子树的结果后，判断如果都有值，则root就是最近公共祖先，一直把该值返回到顶层即可

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function(root, p, q) {
    if(!root || root === q || root === p) {
        return root
    }
    let left = lowestCommonAncestor(root.left, p, q)
    let right = lowestCommonAncestor(root.right, p, q)
    if(left && right) {
        return root
    } else if(!left && right) {
        return right
    } else if(left && !right) {
        return left
    } else {
        return null
    }
};